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IN   MEMORIAM 
FLOR1AN  CAJORI 


A.   StnUtr' 


HAWNEY'S 

COMPLETE  MEASURER: 


OR,    THE 


WHOLE  ART  OF  MEASURING 


A   PLAIN   AND    COMPREHENSIVE 

TREATISE  ON  PRACTICAL  GEOMETRY  AND 
MENSURATION. 

Corrected  and  improved  by  T.  Keith. 
THIRD  EDITION. 


WITH  AX  APPENDIX  CONTAINING  RULES  AND  EXAMPLES  FOR    tflND- 

LSTG   THE  WEIGHT  AND  DIMENSIONS  OF  BALLS  AND   SHELLS, 

WITH  THEIR  APPROPRIATE  QUANTITIES  OF  POWDER. 


REVISED    AND    CORRECTED    BY    JOHN    D.    CRAIGj 
PROFESSOR  OF  MATHEMATICS, 


BALTIMORE : 

PUBLISHED   BY  F.  LUCAS,  JR.  AND  NEAL,  WILLS  &  COLE, 
S.  P.  CHILD  &  CO.  PRINTERS. 

1813. 


\w*\  v^ 

\S'I3 

<: 


.'  pjsti-ici  ?f  >W>lanclj  to  wit : 

*••'••••*  BE  II'  REMEMBERED,  that  on  this  first  day  of 
*  seal.  *  May,  in  the  thirty-seventh  year  of  the  Independence 

********  of  the  United  States  of  America,  FjgJidjngLiicas,  junr. 
of  the  said  district,  liath  deposited  in  this  office,  tli'e  title  of  a 
book  the  right  whereof  he  claims  as  proprietor,  in  the  words 
and  figures  following-,  to  wit : 

"  Hawncy's  Complete  Measurer  :  or,  the  whole  art  of  meas- 
uring. Being  a  plain  and  comprehensive  treatise  on 
"practical  geometry  and  mensuration.  Corrected  and 
"improved  by  T.  Keith.  Third  edition,  with  an  appen- 
dix containing-  rules  and  examples  for  finding-  the  weight 
"and  dimensions  of  balls  and  shells,  with  their  appropri- 
ate qviantities  of  powder.  Revised  and  corrected  by 
"  John  D.  Craig-,  professor  of  mathematics." 

In  conformity  to  the  act  of  the  congress  of  the  United  States, 
intitled,  "  An  act  for  the  encouragement  of  learning  by  securing 
the  copies  of  maps,  charts,  and  books,  to  the  authors  and  pro- 
prietors of  such  copies  during  the  times  therein  mentioned." 
And  also  to  the  act,  intitled  "  An  act  supplementary  to  the  act, 
intitled,  'An  act  for  the  encouragement  of  learning,  by  secur- 
ing the  copies  of  maps,  charts,  and  books,  to  the  authors  and 
proprietors  of  such  copies  during  the  times  therein  mentioned,' 
and  extending  the  benefits  thereof  to  the  arts  of  designing, 
engraving,  and  etching  historical  and  other  prints." 

PHILIP  MOORE, 
t'krk  of  the  District  of  Maryland. 

cajor: 


PREFACE 

BY  THE  EDITOR. 


THE  many  editions  which  Mr.  Hawney's 
Treatise  on  Mensuration  has  gone  through, 
are  evident  proofs  of  its  general  utility.  But 
at  the  time  when  this  work  was  first  pub- 
lished, it  was  not  the  practice  in  schools  for 
each  scholar  to  have  a  printed  book  by  him, 
while  engaged  in  the  study  of  arithmetic  and 
mensuration;  consequently  Mr.  Hawney 
gave  no  particular  examples  as  exercises  for 
a  learner,  without  working  them  at  full  length, 
and  explaining  every  step.  Though  this  was 
perhaps  originally  an  advantage  to  the  book, 
it  precluded  the  use  of  it  in  our  modern 
schools;  for  with  such  assistance,  a  boy  of 
good  abilities  would  naturally  become  indo- 
lent, for  want  of  something  to  exert  Ins  ge- 
nius ;  and  a  boy  of  a  heavy  disposition  would 
be  induced  to  copy  all  his  work  from  the 
book. 

To  remedy  these  inconveniences,  the  pro- 
prietors of  the  work  engaged  the  present  edi- 
tor to  make  such  alterations  and  additions  as 


IV  PREFACE. 

would  render  it  useful  in  schools,  without 
diminishing  its  former  plainness  and  perspi- 
cuity. 

On  examination  of  the  sixteenth  edition, 
the  editor  found  it  to  he  replete  with  errors, 
owing  to  the  inattention,  or  incapacity  of  those 
who  had  the  care  of  the  press,  since  the  death 
of  the  author.  Hence  arose  the  necessity  of 
working  every  example  anew,  and  examining 
every  rule  with  the  greatest  care.  This  was 
attended  with  considerable  labour,  and  at  the 
same  time  was  not  a  very  agreeable  task  ; 
for  as  the  nature  and  plan  of  the  work  has 
undergone  little  or  no  change,  so  the  merit  is 
still  due  to  the  author,  while  the  errors,  if  any 
of  consequence  are  left,  will  be  attached  to 
the  present  editor. 

The  first  nine  chapters  are  in  substance  the 
same  as  in  former  editions.  The  Xth,  Xlth, 
and  XII th  are  added  by  the  editor,  besides 
various  rules  and  observations  throughout  the 
work,  the  principal  of  which  are  distinguish- 
ed in  the  table  of  contents  by  prefixing  an 
asterisk.  - 

The  mensuration  of  the  five  regular  bodies 
is  placed  immediately  after  the  mensuration 
of  solids,  and  these  are  succeeded  by  board 
and  timber  measure. 


PREFACE.  V 

In  the  former  editions,  timber  measuring 
was  divided  into  several  sections  ;  as,  squar- 
ed timber,  or  pieces  of  timber  in  the  form  of 
a  parallelopipedon  ;  unequal  squared  timber, 
or  pieces  of  timber  in  the  form  of  the  frustum 
of  a  pyramid ;  round  timber  with  equal  bases, 
or  pieces  of  timber  in  the  form  of  a  cylinder ; 
round  timber  tviih  unequal  bases,  or  pieces 
of  timber  in  the  form  of  the  frustum  of  a 
cone  :  and  the  contents  were  found,  both  by 
the  customary  method  of  measuring  timber, 
and  likewise  by  the  rule  for  each  respective 
solid.  In  this  edition  these  examples  are 
placed  under  the  several  rules  to  which  they 
belong,  and  are  distinguished  by  asterisks*  ; 
and  in  timber  measuring  they  are  brought 
forward,  and  solved  by  the  customary  method, 
with  reference  to  the  pages  where  they  are 
truly  solved.  By  this  means  several  pages 
have  been  saved  for  the  introduction  of  more 
useful  matter. 

The  gauging  and  surveying,  which  were 
given  in  the  former  editions  by  way  of  ap- 
pendix, are  introduced  in  this  edition  with- 
out that  distinction.  The  former  has  under- 
gone such  alterations  as  will  render  it  very 
useful  for  those  who  are  candidates  for  the 
excise,  or  who  want  to  find  the  contents  of 
different  kinds  of  vessels.     And  the  latter 

-     % 


M  OlIEFACE. 

will  be  found  to  contain  sufficient  information 
for  such  as  want  to  find  the  true  content  of 
any  single  field,  or  parcel  of  land,  by  the 
chain  and  cross-staff  only.  To  form  the 
complete  gauge*  and  surveyer,  recourse  must 
be  had  to  treatises  written  expressly  for  the 
purpose. 

The  practical  questions,  which  proceeded 
the  appendix,  are  now  placed  at  the  end 
of  the  book  and  in  the  same  order  in  which 
they  were  formerly  arranged. 

The  rest  of  the  practical  examples  through- 
out the  work,  which  amount  to  about  four 
hundred,  are  given  by  the  editor ;  several  of 
which  were  copied  from  a  manuscript,  con- 
taining notes  and  observations  on  Hawney, 
by  Mr.  Rawes,  master  of  the  academy  at 
Bromley,  in  Kent. 

In  this  edition,  eighty  new  geometrical  fig- 
ures, illustrating  the  different  problems,  are 
likewise  given  ;  so  that  neither  pains  nor  ex- 
pence  have  been  spared,  to  render  this  work 
of  equal  utility  with  any  other  of  a  similar 
nature. 

Heddon- Court,  Swallow -street, 
September,  1798. 


CONTENTS. 


PART  I. 

ARITHMETIC  AND  GEOMETRY. 
Chap.  Page 

I.  Notation  of  Decimals, 1 

II.  Reduction  of  Decimals,  ....  3 

III.  Addition  of  Decimals, 10 

IV.  Subtraction  of  Decimals,  .  .  .  11 

V.  Multiplication  of  Decimals,  .  .  .12 

VI.  Division  of  Decimals,  .  .  .17 
VII.  Extraction  of  the  Square  Root,             .             .         .25 

VIII.  Extraction  of  the  Cube  Root,         .  .  .  31 

IX.  Duodecimals,  .  .  .  .  .40 

X.  *Of  Gunter's  Scale,  ...  .  .48 

*0*  the  Diagonal  Scale,  ...  51 

XI.  *Of  the  Cahfenteh's  Rule,  .  .  .52 

XII.  *Piiactical  Geometry,  .  .  .  .57 


PART  II. 

CIIAFTEH   I, 

^Mensuration  of  Superficies. 

1.  To  find  the  area  of  a  square,  .  .  79 

2.  To  find  the  area  of  a  rectangled  parallelogram,     80 

3.  To  find  the  area  of  a  rhombus,         ...       82 

4.  To  find  the  area  of  a  rhomboides,       .         .         .83 

5.  To  find  the  area  of  a  triangle,        .         .         .     .   84 
*T\vo  sides  of  a  right-angled  triangle  given  to 

find  the  third,         ....  89 

6.  To  find  the  area  of  a  trapezium,  .  .91 


VJii  CONTEXTS 

rvr.r 

Skit.      *  To  find  the  area  of*  a  trapezoid,  93 

7-  To  find  tin-  ana  of  an  irregular  figure,     .  .      vi 

8.  To  find  the  area  of  a  regular  Polygon,  .        96 

y.    Of  a  Circle,              .  101 

10.  To  find  the  area  of  a  semicircle,        .  .        115 

11»  To  find  the  area  of  a  quadrant,        .        .  .     116 

To  find  the  length  of  any  arch  of  a  circle,  117 

To  find  the   diameter  of  a  circle,  having-  the 

chord  and  versed  sine  given,          .  .         119 

\2-  To  find  the  area  of  the  sector  of  a  circle,  .     120 

13.  To  find  the  area  of  the  segment  of  a  circle,      .     122 

14.  To  find  the  area  of  the  compound  figures,  .     125 

15.  To  find  the  area  of  an  ellipsis,            .          .  '     126 

*  To  find  the  area  of  an  elliptical  segment,  .     128 

*  To  find  the  circumference  of  an  ellipsis,  .     129 

*  To  find  the  area  of  an  elliptical  ring-,        .  .     130 

16.  To  find  the  area  of  a  parabola,          .  .     132 


cnAr-TEE  ii. 


Mensuration  of  Solids. 

Sect.    1.  To  find  the  solidity  of  a  cube,  .         .         .  135 

2.  To  find  the  solidity  of  a  parallelopipedon,         .  137 

3.  To  find  the  solidity  of  a  prism,         .         .         .  139 

4.  To  find  the  solidity  of  a  pyramid,     .         .         .  142 
To  find  the  superficial  content  of  a  pyramid     .  143 

*  The  perpendicular  height  of  a  pyramid  given 

to  find  the  slant  height,  and  the  contrary,      .  144 

5.  To  find  the  solidity  of  a  cylinder,     .         .         .  147 
To  find  the  superficies  of  a  cylinder,         .         .  149 

6.  To  find  the  solidity  of  a  cone,            .         .         .  150 
To  find  the  superficies  of  a  cone,      .         .         .  151 

7.  To  find  the  solidity  of  a  frustum  of  a  pyramid,  153 
To  find  the  superficies  of  the  frustum  of  a  py- 
ramid,      155 

*  To  find  the  slant  height,  or  perpendicular  height 

of  the  frustum  of  a  pyramid,  .        .        .  155 


CONTENTS.  IX 

PAGE 

Sect.     *  To  find  the  perpendicular  height  of  that  pyramid 

of  which  any  given  frustum  is  the  part,         .  155 

*  To  find  the  area  of  the  front  of  a  circular  arch  161 

8.  To  find  the  solidity  of  the  frustum  of  a  cone,  163 

9.  *  To  find  the  solidity  of  a  wedge,          .         ,  165 

10.  To  find  the  solidity  of  a  prismoid,         .         .  167 
To  find  the  solidity  of  a  cylindroid,         .         .  171 

11.  To  find  the  solidity  of  a  sphere  or  globe,         .  174 
To  find  the  solidity  of  the  segment  of  a  sphere,  179 

*  To  find  the  solidity  of  the  frustum  or  zone  of  a 

sphere, 180 

*  To  find  the  convex  surface  of  any  segment  or 

zone  of  a  sphere,               .             .             .  181 

12.  Tq  find  the  solidity  of  a  spheroid,           .         .  185 

*  To  find  the  solidity  of  the  segment  of  a  sphe- 

roid,           .                     .            ...  187 

*  To  find  the  solidity  of  the  middle  zone  of  a 

spheroid,         .....  190 

13.  To  find  the  solidity  of  a  parabolic  conoid,         .  191 

*  To  find  the  solidity  of  the  frustum  of  a  parabo- 

lic conoid,             193 

14.  To  find  the  solidity  of  a  parabolic  spindle,       .  194 

*  To  find  the  solidity  of  the  middle  frustum  of  a 

parabolic  spindle,              .             .             .  197 

*  To  find  the  solidity  of  the  middle  frustum  of 

any  spindle, 198 

15.  To  find  the  solidity  of  the  five  regular  bodies,  199 

16.  To  measure  any  irregular  solid,        .         .         .  203 


CHAPTEIt    III. 

The  Mensuration  of  Board  and  Timber. 

Sect.    1.  To  find  the  superficial  content  of  a  board  or 

plank, 205 

2.  The  customary  method  of  measuring  timber     .     208 

*  A  table  for  measuring  timber,         .         .         .         209 

*  A  general  scholium,  or  remarks,  on  timber  mea- 

suring,          215 


•  "ON'TRNTS. 


nnrrr.n    iv. 


77a'  Mensuration  of  Artificer*'  Work. 

Sbct.    1.  Carpenters'  and  .Toiners'  work,        .        .  '221 

2.  Bricklayers'  work, 229 

3.  Plasterers' work,  238 

4.  Painters'  work,         ....  240 

5.  Glaziers'  work, 242 

6.  Mason's  work,  ....  244 

7.  *  Paviors'  work,  ....  245 


CHAPTER   V. 

Of  Gauging. 

Pkob.     *  Of  the  sliding  rule, 247 

1.  To  find  multipliers,  divisors,  and  gauge-points,  250 

2.  To  find  the  area  in  gallons,  of  any  rectilineal 

plane  figure,  ....  252 

3.  To  find  the  area  of  a  circle  in  ale  gallons,  &c.  254 

4.  To  find  the  area  of  an  ellipsis  in  ale  gallons,  &c,  255 

5.  To  find  the  content  of  any  prism  in  ale  gallons,  256 

6.  To  find  the  content  of  any  vessel  whose  ends 

are  squares,  or  rectangles  of  any  dimensions,  257 

7.  To  find  the  content  in  ale  gallons,  8cc.  of  the 

frustum  of  a  cone, 259 

8.*To  gauge  and  inch  a  tun  in  the  form  of  the  frus- 
tum of  a  cone,             .....  261 
9.  To  gauge  a  copper, 263 

10.  To  compute  the  content  of  any  close  cask,       .  265 

*  To  find  the  content  of  am  cask,        .         .         .  269 

*  A  general  table  for  finding  the  content  of  any 

cask  by  the  sliding  rule,         ....  270 

11.  Of  the  ullage  of  casks,  ....  272 

*  A  table  of  the  areas  of  the  segments  of  a  circle,  278 


CONTENTS.  XI 


CHAPTER   VI. 

Of  Surveying. 

Prob.     *  Of  the  surveying-cross,  or  cross-staff,  .        283 

1.  *  To  measure  off-setts  with  a  chain  -and  cross- 

staff,  284 

2.  To  measure  a  field  in  the  form  of  a  trapezium,    287 

3.  *  To  measure  a  four-sided  field  with  crooked 

hedges, 290 

4.  To  measure  an  irregular  field,         .         .        .       291 

5.  *  To  cut  off  from  a  plan  a  given  number  of  a- 

cres, 295 


CHAPTER    VII. 

Practical  questions  in  measuring,         ....        295 
Explanation  of  characters  made  use  of  in  the  work,       .    308 

APPENDIX. 

Of  the  weight  and  dimensions  of  Balls  and  Shells. 


THE 


COMPLETE  MEASURES. 


PAIIT  I. 


CHAPTER  I. 

Notation  of  Decimals, 

A.  DECIMAL  fraction  is  an  artificial  way  of  setting 
down  and  expressing  natural,  or  vulgar  fractions,  as 
whole  numbers.  A  decimal  fraction  has  always  for 
its  denominator  an  unit,  with  a  cypher  or  cyphers  an- 
nexed to  it,  and  must  therefore  be  either  10,  100, 
1000,  10000,  &c.  and  consequently  in  writing  down  a 
decimal  fraction  there  is  no  necessity  for  writing  down 
the  denominator :  as  by  bare  inspection,  it  is  certainly 
known,  consisting  of  an  unit  with  as  many  cyphers 
annexed  to  it  as  there  are  places  (or  figures)  in  the 
numerator. 

Examples.  The  decimal  fraction  T**y  may  be  writ- 
ten thus,  .25,  its  denominator  being  known  to  be  an 
unit  with  two  cyphers  ;  because  there  are  two  figures 
in  the  numerator.  In  like  manner,  ^**v  may  be  thus 
written,  .125:    ^gfo  thus,   .3575  ;    -Jf^  thus,   1078} 

and  iHtns  thus>  -00*5' 

B 


lotion  of  Decimals. 

\.i  -aIwIi  lei i  rease  in  a  ten-fold  proportion, 

tOwardl  the  li -I't  hand,  si)  on  the  contrary,  decimals  de- 
crease towards  the  right  hand  in  the  same  proportion. 
as  in  the  following  table. 


A 


c 

Xi 

.3      H 


a  _g  li    — ! 

5  -3  ,:  -  9 


~  2  «4 

K    f     ^     Oj        „ 


3  PlbTl 


Z    eg  I  afi^I^I 

76543210, 123456 


Hence  it  appears,  that  cyphers  put  on  the  right 
hand  of  whole  numbers,  increase  the  value  of  those 
numbers  in  a  ten-fold  proportion :  But  being  annexed 
to  the  right  hand  of  a  decimal  fraction,  neither  in- 
crease nor  decrease  the  value  of  it :  So  T2^°^y  is  equi- 
valent to  T2^  or  .25.  And,  on  the  contrary,  though  in 
whole  numbers,  cyphers  before  them,  neither  increase 
nor  diminish  the  value  $  yet  cyphers  before  a  decimal 
fraction  diminish  its  value  in  a  ten-fold  proportion  : 
For  .25,  if  you  put  a  cypher  before  it,  becomes  -ft2^  or 
.025:  And  .123  is  ■fgjgfo  by  prefixing  two  cyphers 
thus,  .00125.  And  therefore  when  you  are  to  write  a 
decimal  fraction,  whose  denominator  has  more  cyphers 
than  there  are  figures  in  the  numerator,  the  places  of 
such  figures  must  be  supplied  by  placing  cyphers  be- 
fore the  figures  of  your  numerator ;  as,  suppose  -j^^ 
were  to  be  written  down,  without  its  denominator; 
here,  because  there  are  three  cyphers  in  the  denomi- 
nator, and  but  two  figures  in  the  numerator,  therefore 
put  a  cypher  before  19,  and  set  it  down  thus,  .010. 


Reduction  of  Decimals. 


CHAPTER  II. 


Reduction  of  Decimals. 

IN  Reduction  of  Decimals,  there  are  three  eases  : 
1st,  To  reduce  a  vulgar  fraction  to  a  decimal.  2dly, 
To  find  the  value  of  a  decimal  in  the  known  parts  of 
coin,  weights,  measures,  &c.  and  3dly,  To  reduce  coin, 
weights,  measures,  &c.  to  a  decimal. 

I.  To  reduce  a  Vulgar  Fraction  to  a  Decimal. 

THE    RULE. 

As  the  denominator  of  the  given  fraction  is  to  its 
numerator,  so  is  an  unit  (with  a  competent  number 
of  cyphers  annexed)  to  the  decimal  required. 

Therefore,  if  to  the  numerator  given,  you  annex  a 
competent  number  of  cyphers,  and  divide  the  result  by 
the  denominator,  the  quotient  is  the  decimal  equivalent 
to  the  vulgar  fraction  given. 

Example  1.  Let  |  be  given,  to  be  reduced  to  a 
decimal  of  two  places,  or  having  100  for  its  denomi- 
nator. 

To  3  (the  numerator  given)  annex  two  cyphers,  and 
it  makes  300,  which  divide  by  the  denominator  1,  and 
the  quotient  is  .75,  the  decimal  required,  and  is  equi- 
valent to  1  given. 

Note.  That  so  many  eyphers  as  you  annex  to  the 
given  numerator,  so  many  places  must  be  pointed  oft* 
in  the  decimal  found ;  and  if  it  should  happen,  that 
there  are  not  so  many  places  of  figures  in  the  quoti- 
ent, the  deficiency  must  be  supplied,  by  prefixing  cy- 
phers to  the  quotient  figures,  as  in  the  next  example. 


.!  Reduction  of  Decimal*- 

Example  2.  Let  -^  be  reduced  to  a  decimal  hav- 
ing six  places. 

To  tlie  numerator  annex  six  cyphers,  and  divide 
by  the  denominator,  and  the  quotient  is  5235,  hut  it 
was  required  to  have  six  places,  therefore  you  must 
put  two  cyphers  before  it,  and  then  it  will  be  .005235, 
which  is  the  decimal  required,  and  is  equivalent 
ro^. 

See  the  work  of  these  two  examples. 

4)3.G0(.75  573)3.000000(.005.233 

28 


20       '  1350 

20  

2040 

3210 

345 

in  the  second  example  there  remains  345,  which  re- 
mainder is  very  insignificant,  it  being  less  than  totto? 
part  of  an  unit,  and  therefore  is  rejected. 

PRACTICAL    EXAMPLES. 

3.  Reduce  -J  to  a  decimal.  Jlns.  .57142  rem.  4, 

4.  Reduce  2fj  to  a  decimal. 

Jns.  .0041152263  rem.  91. 

5.  Reduce  £  of  §  of  A  to  a  decimal. 

Jlns.  .20833,  &c'. 

6.  Reduce  15  fa  to  a  mixed  decimal. 

Jins.  15.38461  rem.  7. 

Jlns.  .17241379  rem.  9. 
8.  Reduce  T^T  to  a  decimal. 

Jlns.  ,026178010471  rem.  39. 


Reduce  ^  to  a  decimal. 


Reduction  of  Decimals.  & 

Note.  A  finite  decimal  is  that  which  ends  at  a 
certain  number  of  places,  such  for  instance  as  ex- 
ample 1.  But  an  infinite  decimal  is  that  which  no 
where  ends,  but  is  understood  to  be  indefinitely  con- 
tinued, such  as  example  3. — In  short,  all  fractions 
whatever,  whose  denominators  are  not  composed  of  2" 
or  5,  or  both,  will  have  their  correspondent  decimal 
infinite.  The  method  of  managing  circulating  deci- 
mals  may  be  met  with  in  KeiWs  Arithmetic,  and  se- 
veral others ;  but  for  common  use,  all  the  decimals, 
beyond  three  or  four  places,  may  be  safely  rejected, 
without  affecting  the  truth  of  the  conclusion. 

II.  To  find  the  value  of  a  Decimal  in  the  known  part? 
of  money,  weight,  measure,  <$*c. 

THE  RULE. 

Multiply  the  given  decimal  by  the  number  of  parts 
in  the  next  inferior  denomination,  and  from  the  pro- 
duct point  oft*  so  many  figures  to  the  right  hand  as 
there  were  figures  in  the  decimal  given ;  and  multiply 
those  figures  pointed  oft"  by  the  number  of  parts  in  the 
next  inferior  denomination,  and  point  off  so  many 
places  as  before,  and  thus  continue  to  do  till  you  have 
brought  it  to  the  lowest  denomination  required. 

Example.  1.  Let  .7565  of  a  pound  sterling  be  given 
to  be  reduced  to  shillings,  pence  and  farthings. 

Multiply  by  20,  by  12,  and  by  4,  as  the  rule  directs, 
and  always  point  oft' four  figures  to  the  right  hand,  and 
you  will  find  it  make  15s.  id.  2q.  See  the  work* 
.7563 
?       20 

s. . 

15.1300 
12 

d. 

1.5600 

4  t 

?• 

2.2400 


tf  Reduction  of  Decimals. 

Example  2.  Let  .50755  of  a  pound  troy  be  reduced 
to  ounces,  penny- we it^hts  :ind  grain* 

Multiply  by  12,  by  20,  and  by  21,  and  always  point 
oft*  five  figures  towards  the  right  band,  and  you  will 
find  the  answer  to  be  7  oz.  Sdwts.  i0.gr.  fere.  Bee  the 
Vorfc. 

.59755" 
12 


,17000 

20  oz.  pivts.     pr. 
FacifT        3     9.8SS 


3.41200 
24 

164800 
82400 

9.88800 


Example  3.  Let  .43569  of  a  ton  be  reduced  fo 
hundreds,  quarters,  and  pounds. 

Multiply  by  20,  by  4,  and  by  28,  and  the  answer 
will  be  8  C.  2  qrs.  24  lb.  fere. 


.43569 
20 

8.71380 

4  C.   qrs.    lb. 
Facit     8       2     23.9456 


2.85520 


23.94560 


Reduction  of  Decimals*  ? 

Example.  4.  Let  .9595  of  a  foot  be  reduced  info 
inches  and  quarters. 

.9595 
12 


11.5140 

4      Facit  11  inches  2  quarters. 


3.0560 


PRACTICAL    EXAMPLES. 

5.  What  is  the  value  of  .7575  of  a  pound  sterling  ? 

J&ns.  15s.  lfd.  .2, 

6.  Required  the  value  of  .75435  of  a  shilling  ? 

Jins.  9.0522  pence, 

7.  What  is  the  value  of  .375  of  a  guinea  ? 

Jlns.  7s.  10^,d. 

8.  What  is  the  value  of  .4575  of  a  hundred  weight? 

Jlns.  1  qr.  23  lb.  3  oz.  13.44  drams. 

9.  What  is  the  value  of  .175  of  a  ton  avoirdupois  t5 

Jlns.  3  cwt.  2  qrs. 

10.  Required  the  value  of  .02575  of  a  pound  troy  ? 

Jins.  6  dwt.  4.32  grs. 

11.  What  is  the  value  of  .04535  of  a  mile  ? 

Jlns.  14 p.  2yds.  2ft.  5  in.  1.128  barley-corn, 

12.  What  is  the  value  of  .6375  of  an  acre  ? 

Jlns.  2  roods  22  perches. 

13.  What   is  the  value  of  .574  of.  a  hogshead  of 
beer  ?  Jlns.  30  gal.  3  qt.  1.968  pt. 

14.  What  is  the  value  of  .4285  of  a  year  ? 

Jlns.  156  days,  12hrs.  13  m.  51  sec.  36  thirds. 

III.  To  reduce  the  known  parts  of  money,  weights,  mea* 
sure,  Sfc.  to  a  decimal. 

THE  RULE. 

To  the  number  of  parts  of  the  less  denomination 
given,  annex  a  competent  number  of  cyphers,  and 


8  Reduction  of  Decimals. 

divide  by  the  number  of  Mich  parts  that  are  contain- 
ed in  the  greater  denomination,  to  which  the  deei- 
m;il  is  to  be  brought;  and  the  quotient  is  the  decimal 
sought. 

Example  1.  Let  Qd.  be  reduced  to  the  decimal  of  a 
pound. 

To  G  annex  a  competent  number  of  cyphers  (sup- 
pose 3,)  and  divide  the  result  by  210  (the  pence  in  a 
pound,)  and  the  quotient  is  the  decimal  required. 

240)6.000(.023 


1200 

Facit  .023 


Example  2.  Let  3c?.  |  be  reduced  to  the  decimal  of 
a  pound,  having  six  places. 

In  3d.  %  there  are  fifteen  farthings,  therefore  to  13 
annex  six  cyphers  (because  there  are  to  be  six  places 
in  the  decimal  required,)  and  divide  by  960  (the  far- 
things in  a  pound,)  and  the  quotient  is  .013625. 

9610)13.00000l0(.013625 


540 
600 
240 
480 


Example.  3.  Let  3  \  inches  he  redueed  to  the  deci- 
mal of  a  foot,  consisting  of  four  places. 


Reduction  of  Decimals.  9 

In  3£  inches,  there  are  13  quarters ;  therefore  fo  13 
annex  four  cyphers,  and  divide  by  48  (the  quarters  in 
a  foot)  and  the  quotient  is  .2708. 

48)13.0000(.2708 


340 


400 


16 


Example  4.  Let  9  C.  ±qr.  16/6.  be  reduced  to  the 
decimal  of  a  ton,  having  6  places. 

C.    qr.     Ik 

9        1        16  2240)1052.000000(.469642 


** 

37  qrs. 

28 

15600 

302 

75 

21600 
14400 

1052  Pounds 

9600 

6400 

1920 
Facit  .469642. 

PRACTICAL    EXAMPLES. 

5.  Reduce  7s.  5|d.  to  the  decimal  of  a  pound  r 

Ans.  1. 3729166,  &c. 

6.  What  decimal  part  of  a  pound  sterling  is  three- 
halfpence  ?  Jins.  1 .00625. 

7.  Reduce  4s.  719Td.  to  the  decimal  of  a  pound  ster- 
ling i  Ms,  1 .2325757,  &c 


10  Addition  of  Decimals. 

8.  Reduce  10  o%,  Ji  dwt.  3 gr.  to  the  decimal  of  a 
pound  Ti  Ans.  ,1296870  ft. 

9.  Reduce  Scui.    1  qr.    1 1  lb.   to   the  decimal   of  a 
Ton?  Ans.  .16870  Ton. 

10.  Reduce  22  feet   7   inches  to  the  decimal   of  a 
Foot  ?  Ans.  22.08*8  Feet 

11.  Reduce  2qrs.  15  lb.  to  the  decimal  of  a  hundred 
weight?  Ans.  .6336285711,  &c.  cwt. 

12.  What  decimal  part  of  a  year  is  3w.  4?d.  55  hours, 
reckoning  363  days  6  hours  a  Year  ? 

Ans.   .074720511065  yr. 

13.  Reduce  2.15  shillings  to  the  decimal  of  a  pound  ? 

Ans.  1 .1225. 

14.  Reduce  1.074  roods  to  the  decimal  of  an  Acre  ? 

Ms,  .2685  Acre. 
13.  Reduce  17.69  yards  to  the  decimal  of  a  mile  ? 

Ms.  .010031136  m. 


CHAPTER  m. 

Addition  of  Decimals. 

ADDITION  of  decimals  is  performed  the  same  way 
as  Addition  of  whole  numbers,  only  yon  must  observe 
to  place  your  numbers  right,  that  is  all  the  decimal 
points  under  each  other,  units  under  units,  tenths  un- 
der tenths,  &c. 

Example.  Let  317.25;  17.125;  275.5;  47.3579:  and 
12.75  ;  be  added  together  into  one  sum. 

317.25 

17.125 
275.5 

47.3079 

12.75 


Sum  669.9825 


Subtraction  of  Decimals.  11 

PRACTICAL    EXAMPLES. 

2.  Add  5.714;  3.456;  .543;  17.4957  together. 

Sum  27.2087. 

3.  Add  3.754;  47.5;  .00857;  and  37.5  together. 

Sum  S8.7G257. 
1.  Add  54.34;  .375;  14.795;  and  1.5  together. 

Sum  71.01, 


CHAPTER  IV. 


SuB'TRACflON  Of  DECIMALS. 

SUBTRACTION  of  decimals  is  likewise  perform- 
ed the  same  way  as  in  whole  numbers,  respect  being 
had  (as  in  addition)  to  the  right  placing  of  the  num- 
bers, as  in  the  following  examples. 

(1)  (2) 

From  212.0137  From  201.125 

Subtr.    31.1275  Subtr.      5.57846 


Rests   180.8862  Rests  195.54634 


Note.  If  the  number  of  places  in  the  decimals  be 
more  in  that  which  is  to  be  subtracted,  than  in  that 
which  you  subtract  from,  you  must  suppose  cyphers 
to  make  up  the  number  of  places  as  in  the  second  Ex- 
ample. 


PRACTICAL    EXAMPLES. 


3.  Required  the  difference  between  57.49  and  5.768  ? 

Jins.  51.722, 


12  Multiplication  of  Decimals. 

4.  "What  is  the  difference  between  .3054  and  3.051  ? 

Jlns.  i.risG. 

5.  Required     (lie    difference   between    1745.3    and 
173.45?  4n$.  1.371.85. 

6.  What  is  (he  difference  between  seven  tenths  of  an 
unit,  and  fifty-four  ten  thousand  parts  of  an  unit? 

Jlns.  .6940. 

7.  What  is  the  difference  between  .105  and  1.00075? 

Jlns.  .89575. 

8.  What  is  the  difference  between  150.43  and  755.355? 

Jlns.  604.925. 

9.  From  1754.754  take  375,494478  ? 

Jlns.  1379.259522. 

10.  Required   the   difference  between   17.541    and 
35.49  ?  Jlns.  17.949. 


CHAPTER  V. 

Multiplication  of  Decimals. 

MULTIPLICATION  of  decimals  is  also  performed 
the  same  way  as  Multiplication  of  whole  numbers  ; 
but  to  know  the  value  of  the  product,  observe  this 
Rule  : 

Cut  off,  or  separate  by  a  comma  or  point,  so  many 
decimal  places  in  the  product,  as  there  are  places  of 
decimals  in  both  factors,  viz.  both  in  the  multiplicand 
and  multiplier. 

Example  1.  Let  3.125  be  multiplied  by  2.75. 
Multiply  the  numbers  together,  as  if  they  were  whole 
numbers,  and  the  product  is  8.59375  :  And  because 
there  are  three  places  of  decimals  pointed  off  in  the 
multiplicand,  and  two  places  in  the  multiplier,  there- 
fore you  must  point  off  five  places  of  decimals  in  the 
product,  as  you  may  see  by  the  work. 


Multijilication  of  Decimals  * 

Multiplicand  3.125 
Multiplier         2.75 


15625 

21875 
6250 


Product  8.59375 

Example  2.  Let  79.25  be  multiplied  by  .439. 
In  this  example,  because  two  places  of  decimals  are 
pointed  oft'  in  the  multiplicand,  and  three  in  the  mul- 
tiplier, therefore  there  must  be  five  pointed  off  in  the- 
product. 

Multiplicand  79.25 
Multiplier         .459 

71325 

39625 
31700 

Product  36.37375 

Example  3.  Let  1.35272  be  multiplied  by  .00423. 
In  this  example,  because  in  the  multiplicand  arc 
six  decimal  places  and  in  the  multiplier  live  places ; 
therefore  in  the  product  there  must  be  eleven  places 
of  decimals  $  but  when  the  multiplication  is  finished, 
the  product  is  but  57490600  viz.  only  eight  places  ; 
therefore,  in  this  case,  you  must  put  three  cyphers  be- 
fore the  product  figures,  to  make  up  the  number  of 
eleven  places:  So  the  true  product  will  be  .00057490600. 

Multiplicand  .135272 
Multiplier  .00425 

676360 
270544 
541088 


Product  .00057490600 


U  tracted  Multiplicati 

PRACTICAL    K: 

i.  Multiply  .001 172  by  .1013. 

Product  .0001538240, 
-».  Multiply  4)17533  by  3*7.        Product  6.083604. 
6.  Multiply  2T9.23  by  .445.        Product  I2t.j> 
Multiply  32.07.32  by  .032  0. 

Product  1.0.1211  MM). 
8.  Multiply  1. 1 13  bv  15.98.  Product  70.99914. 

Multiply  20.0291  by  35.45. 

Product  710.031595. 

10.  Multiply  7.3564  by  .0126. 

Product  .09269064. 

11.  Multiply  .75432  by  .0356. 

Product  .026853792. 

12.  Multiply  .004735  by  .0374. 

Product  .0001770890. 


Contracted  Multip lic j r jos  of  Decimals. 

Because  in  multiplication  of  decimal  parts,  and  mix- 
ed numbers,  there  is  no  need  to  express  all  the  figures 
of  the  product,  but  in  most  cases  two,  three  or  four 
places  of  decimals  will  be  sufficient;  therefore,  to  con- 
tract the  work,  observe  the  following 


Write  the  unit's  place  of  the  multiplier  under  that 
place  of  the  multiplicand,  which  you  intend  to  keep 
in  the  product:  then  invert  the  order  of  all  the  other 
figures;  that  is  write  them  all  the  contrary  way;  and 
in  multiplying,  begin  always  at  that  figure  in  the  mul- 
tiplicand which  stands  over  the  iigure  you  are  then 
multiplying  withal,  and  set  down  the  first  figure  of 
each  particular  product  directly  one  under  the  other: 
But  take  care  to  increase  the  first  figure  of  every  Hue 


Contracted  Multiplication.  15 

of  the  product,  with  what  Mould  arise  by  carrying  1 
from  5  to  15 ;  2  from  15  to  25  ;  3  from  25  to  35,  &c. 
from  the  product  of  the  two  figures  (in  the  multipli- 
cand) on  the  right-hand  of  the  multiplying  figure. 

Example  1.  Let  2.38645  be  multiplied  by  S.2175, 
and  let  there  be  only  four  places  retained  in  the  de- 
cimals of  the  product. 

First,  according  to  the  directions,  write  down  the 
multiplicand,  and  under  it  write  the  multiplier,  thus  : 
place  the  8  (being  the  unit's  place  of  the  multiplier) 
under  4,  the  fourth  place  of  decimals  in  the  multipli- 
cand, and  write  the  rest  of  the  figures  quite  contrary 
to  the  usual  way,  as  in  the  following  work :  Then 
begin  to  multiply,  first  the  5  which  is  left  out,  only 
with  regard  to  the  increase,  which  must  be  carried 
from  it  i  saying,  8  times  5  is  40 ;  carry  four  in  your 
mind,  and  say,  8  times  4  is  32,  and  4  I  carry,  is  36  ; 
set  down  6,  and  carry  3,  and  proceed  through  the  rest 
of  the  figures  as  in  common  multiplication.  Then 
begin  to  multiply  with  2;  saying,  2  times  5  is  10, 
nought  and  carry  1;  2  times  4  is  8,  and  1  is  9, 
for  which  I  carry  1,  because  it  is  above  5;  then  2 
times  6  is  12,  and  1  that  I  carry  is  13;  set  down  3  and 
carry  1 ;  and  proceed  through  the  rest  of  the  figures 
as  in  common  multiplication.  Then  multiply  with  1  : 
saying,  once  6  is  6,  for  which  I  carry  1,  and  say,  onee 
8  is  8,  and  1  is  9 ;  set  down  9,  and  proceed  as  in  com- 
mon multiplication.  Then  multiply  with  7:  saying, 
7  times  6  is  42,  2  and  carry  4;  7  times  S  is  56,  and  4 
is  60,  nought  and  carry  6 ;  7  times  3  is  21,  and  6  is  27  ; 
set  down  7  and  carry  2,  and  proceed.  Lastly,  multi- 
ply with  5 :  saying,  5  times  8  is  40,  nought  and  carry 
4;  5  times  3  is  15,  and  4  is  19 ;  for  which  carry  2,  and 
say,  5  times  2  is  10,  and  2  that  I  carry  is  12:  which 
set  down,  and  add  all  the  products  together,  and  the 
total  will  be  19.6107. — See  the  work. 


16 


Contracted  Multiplication. 


Contracted. 

Common. 

9«#8 

8.  2 173 

100916 

11 

167 

0513 

615 

167 

90 

12 

190916 

0 

1  9.6 107 

19.6106  |  52873 

I  kave  here  set  down  the  work  of  the  last  example, 
w  rooght  by  the  common  way,  by  which  you  may  see 
the  reason  of  the  contracted  way,  all  the  figures  on 
•he  right-hand  of  the  line  being  wholly  omitted. 

Example  3.  Let  375.13758  be  multiplied  by  16.7324, 
so  that  the  product  may  have  but  four  places  of  de 
vinials. 

375.13753  the  -Multiplicand. 
4237.61  the  Multiplier  reversed. 

513758  the  product  with  1 
22508255  the  prod,  with  6  increased  with  6X8 
5963  the  prod,  with  7  increased  with  7x^X3 
112511  the  prod,  with  8  increased  with  3X5X7 
7503  the  prod,  with  2  increased  with  2X7X3 
1500  the  prod,  with  4  increased  with  o 

6276.9520  the  product  required. 


PRACTICAL    EXAMPLES. 


Example  3.  Multiply  395.3756  by  .75642. 

Ans.  299.0699. 


Division  of  Decimals.  17 

4.  Let  54.7494367  be  multiplied  by  4.724733  reserv- 
ing only  five  places  of  decimals  in  the  product. 

Jlns.  258.67758. 

5.  Multiply  475.710564  by  .3416494  and  retain  three 
decimals  in  the  product.  Jlns.  162.525. 

6.  Let  4745.679  be  multiplied  by  751.4519,  and  re- 
serve only  the  integers,  or  whole  numbers,  in  the  pro- 
duct. Ms.  3566163; 


CHAPTER  VI. 

Division  of  Decimals. 

DIVISION  of  decimals  is  performed  in  the  same 
manner  as  division  of  whole  numbers:  to  know  the 
value  or  denomination  of  the  quotient,  is  the  only  dif- 
ficulty ;  for  the  resolving  of  which,  observe  either  of 
the  following 

RULES. 

I.  The  first  figure  in  the  quotient  must  be  of  the 
same  denomination  with  that  figure  in  the  dividend 
which  stands  (or  is  supposed  to  stand)  over  the  unit's 
place  of  the  product  of  the  first  quotient  figure  by  the 
divisor. 

II.  When  the  work  of  division  is  ended,  count  how' 
many  places  of  decimal  parts  there  are  in  the  divi- 
dend more  than  in  the  divisor;  for  that  excess  is  the 
number  of  places  which  must  be  separated  in  the  quo- 
tient for  decimals.  But  if  there  be  not  so  many  figures 
in  the  quotient  as  there  are  in  the  said  excess,  that 
deficiency  must  be  supplied,  by  placing  cyphers  before 
the  significant  figures,  towards  the  left-hand,  with  a 
point  before  them;  and  thus  you  will  plainly  discover 
the  value  of  the  quotient. 

c  2 


th  Division  of  Decimals. 

'fliese  following  directions  ought  also  to  be  carefulTtf 
observed. 

If  the  divisor  consist  of  more  places  than  the  divi- 
dend, there  must  be  a  competent  number  of  cyphers 
annexed  to  the  dividend,  to  make  it  consist  of  as  many 
(at  least)  or  more  places  of  decimals  than  the  divisor  3 
for  the  cyphers  added  must  he  reckoned  as  decimals. 

Consider  whether  there  be  as  many  decimal  parts 
in  the  dividend  as  there  are  in  the  divisor;  if  there 
be  not,  make  them  so  many,  or  more,  by  annexing. 
cyphers. 

In  dividing  whole  or  mixed  numbers,  if  there  he  a 
remainder,  you  may  bring  down  more  cyphers;  and, 
by  continuing  your  division,  carry  the  quotient  to  as 
many  places  of  decimals  as  you  please. 


Example  1.  Let  48  be  divided  by  144. 
Tn  this  example  the  divisor  144-  is  greater  than  the 
dividend  48;  therefore,  according  to  the  directions 
above,  I  annex  a  competent  number  of  cyphers  (viz. 
(bur,)  with  a  point  before  them,  and  divide  in  lire 
i*ual  way, 


i44)4S.0000(.333:'i 
432 

430 


lint,  first,  in  seeking  hoir  often  144  in  48.d  (t]\Q 
first  three  figures  of  the  dividend,)  I  find  the  unit's 
place  of  the  product  of  the  first  quotient  figure  by  ther 


Division  of  Decimate.  $9 

divisor  to  fall  under  the  first  place  of  decimals ;  there- 
fore the  first  figure  in  the  quotient  is  in  the  first  place 
of  decimals :  Or,  by  the  second  rule,  there  being  four 
places  of  decimals  in  the  dividend,  and  none  in  the 
divisor ;  so  the  excess  of  decimal  places  in  the  divi- 
dend, above  that  in  the  divisor,  is  four ;  so  that  when 
the  division  is  ended,  there  must  be  four  places  of  de- 
cimals in  the  quotient. 

Example  2.  Let  217.75  be  divided  by  65. 
First,  in  seeking  how  often  65  in  217  (the  first  three 
figures  of  the  dividend)  I  find  the  unit's  place  of  the 
product  of  the  first  quotient  figure  by  the  divisor  to 
fall  under  the  unit's  place  of  the  dividend ;  therefore 
the  first  figure  in  the  quotient  will  be  units,  and  all 
the  rest  decimals:  Or,  by  the  second  rule,,  there  being 
two  places  of  decimals  in  the  dividend,  and  no  deci- 
mals in  the  divisor,  therefore  the  excess  of  decimal 
places  in  the  dividend,  above  the  divisor,  is  two ;  so 
when  the  division  is  ended,  separate  two  places  in  the 
quotient,  towards  the  right-hand  by  a  point, 

65)217.75(3.35 
195.  . 


227 


225 


Example  3.  Let  2F6.15975  be  divided  by  13..2& 


13.25)267.15975(20.163 
2650 


2159 
8347 
3975 


Igj  Div'mon  of  Dcc.lv. 

In  this  third  example  the  unit's  place  of  the  prd- 
duet  of  the  first  quotient  figure  by  the  divisor  falls 
Ullder  6,  the  ten's  place  of  the  dividend;  therefore, 
(by  the  first  rule)  the  first  figure  in  the  quotient  is 
tcus:  Or,  by  the  second  rule,  the  excess  of  decimal 
places  in  the  dividend,  above  the  divisor,  is  three  ; 
there  being  five  places  of  decimals  in  the  dividend, 
and  but  two  in  the  divisor,  so  there  must  be  three  pla- 
ces of  decimals  in  the  quotient. 

Example  4.     Let  13.673139  be  divided  by  sY8t89\ 

;;;.i.s9)i.,5.r,r.jt59(.04ir 

"130355 


03935 


263669 


346 


In  this  fourth  example,  the  unit's  place  of  the  pro- 
duct of  the  first  quotient  figure  by  the  divisor,  falls 
under  7,  the  second  place  of  decimals  in  the  dividend; 
therefore  (by  the  first  rule)  the  first  figure  in  the  quo- 
tient is  in  the  second  place  of  decimals ;  so  that  you 
must  put  a  cypher  before  the  first  figure  in  the  quo- 
tient ;  and  by  the  second  rule,  the  excess  of  decimal 
places  in  the  divisor  is  4;  for  the  decimal  places  in 
the  dividend  are  6,  and  the  number  of  places  in  the 
divisor  but  two ;  therefore  there  must  be  four  places 
of  decimals  in  the  quotient :  But  the  division  being 
finished  after  the  common  way,  the  figures  in  the  quo- 
tient are  but  three,  therefore  you  must  put  the  cypher 
hefore  the  significant  figures. 


Division  of  Decimals.  21 

Example  5.     Let  72.1564  be  divided  by  .134?, 

.134:7)72.1564(535.68 
6735  ,  . 


4806 


'654 


9190 


410S0 
304. 

In  this  example,  the  divisor  being  a  decimal,  the 
last  figure  of  the  product  of  the  first  quotient  figure 
by  the  divisor  falls  under  the  ten's  place  in  the  divi- 
dend, therefore  the  units  (if  there  had  been  any) 
would  fall  under  the  hundreds  place  in  the  dividend, 
and  so  the  first  figure  in  the  quotient  is  hundreds. 
And  by  the  second  rule,  there  being  four  places  of 
decimals  in  the  dividend,  as  many  in  the  divisor,  so 
the  excess  is  nothing;  but  in  dividing  I  put  two  cy- 
phers to  the  remainders,  and  continue  the  division 
to  two  places  further  ;  so  I  have  two  places  of  deci^ 
mals. 

Example  6.     het  .125  be  divided  by  .0457:. 

,0457).1250000(2.735 
0914.  .  . 


3560 


1610 
2390 


Ifl  Division  of  Decimals. 

In  this  example,  the  unit's  place  of  the  product  of 
the  tirst  <| not ii* nt  figure  by  the  divisor  (if  there  had 
any)  Would  fall  under  the  unit's  place  of  the 
dividend :  therefore  the  first  figure  of  the  quotient  IB 
units.  And,  by  the  second  rule,  their  being  seven  pla- 
ces of  decimals  in  the  dividend,  and  but  lour  places 
in  the  divisor,  so  the  excess  is  three  ;  therefore  there 
must  be  three  places  of  decimals  in  the  quotient. 


PRACTICAL  EXAMPLES. 

7.  Divide  .0000059791  by  .00436. 

Quotient  .00131 

8.  Divide  an  unit  by  282,  or,  in  other  words,   find 
the  reciprocal  of  282.  Quotient  .0033-101 

9.  Divide  A  by  .323.  Quotient  1.2307 

10.  Divide  493  by  .012.  Quotient  11783.71 

11.  Divide  .475321  by  97.453.  Quotient  0048774 

12.  Divide  17.543273  by  123.7.  Quotient  .13930 

13.  Divide  143734.35  by  .7493.    Quotient  191851.528 

14.  Divide  16  by  960.  Quotient  .01666,  &c. 
13.  Divide  12  by  1728                 Quotient  .006944,  Sec. 

16.  Divide  47.5493  by  34.75         Quotient  1.36832517 

17.  Divide  70.3571  by  .00573.      Quotient  12976.3062 

18.  Divide  .3754  by  73.714.  Quotient  .004958131 


Division  of  Decimals  contracted. 

In  division  of  decimals  the  common  way,  when  the 
divisor  has  many  figures,  and  it  is  required  to  con- 
tinue the  division  till  the  value  of  the  remainder  be 
but  small,  the  operation  will  sometimes  he  long  and 
tedious,  but  may  be  contracted  by  the  following 
method. 


Division  of  Decimals. 


23 


THE    RULE. 


"By  the  first  rule  of  this  chapter  (page  17,)  find 
v  hat  is  the  value  of  the  first  figure  in  the  quotient : 
then  by  knowing  the  first  figure's  denomination,  you 
may  have  as  many  or  as  few  places  of  decimals  as 
you  please,  by  taking  as  many  of  the  left  hand  figures 
of  the  divisor  as  you  think  convenient  for  the  first  di- 
visor; and  then  take  as  many  figures  of  the  dividend 
as  will  answer  them ;  and,  in  dividing,  omit  one  figure 
of  the  divisor  at  each  following  operation  ;  observing 
to  carry  for  the  increase  of  the  figures  omitted,  as  in 
contracted  multiplication. 

Note.  When  there  are  not  so  many  figures  in  the 
divisor  as  are  required  to  be  in  the  quotient,  begin  the 
division  with  all  the  figures,  as  usual,  and  continue 
dividing  till  the  number  of  figures  in  the  divisor  is 
equal  to  the  number  of  figures  remaining  to  be  found 
in  the  quotient ;  after  which  use  the  contraction. 

Example  1.  Let  721.17562 be  divided  by  2.257432; 
Sclet  there  be  three  places  of  decimals  in  the  quotient. 
Contracted.  Common. 


25743)721.175—62(319.467 
677229 


43946 
22574 


21372 
20317 


1055 
903 


152 
135 


17 
10 


2.25743)721.17562(319.467 


077229 

•* 

43946 

6 

22574 

3 

21372 

3  3 

20316 

87 

1055 

450 

902 

972 





152 

4780 

135 

4458 

17 

03220 

15 

80201 

1 

23019 

|fl  Contracted  Divisit 

Tn  this  example,  the  unit's  place  of  the  product  of 
the  first  quotient  figure  by  the  divisor  falls  under  the 
hundred's  place  in  the  dividend,  and  it  is  required, 
that  three  places  of  decimals  he  in  the  quotient,  so 
there  must  he  six  places  in  all  ;  that  is,  three  places 
of  whole  numbers,  and  three  places  of  decimals. 
Then,  because  1  can  have  the  divisor  in  the  first  six 
figures  of  the  dividend,  I  cut  oft*  the  62  with  a  dash 
of  the  pen,  as  useless;  then  I  seek  how  often  the  divi- 
sor is  in  the  dividend,  and  the  answer  is  three  times  ; 
put  three  in  the  quotient,  and  multiply  and  subtract 
as  in  common  division,  and  the  remainder  is  43940. 
Then  point  oft'  three  in  the  divisor,  and  seek  how  oft- 
en the  remaining  figures  may  be  had  in  43946,  the  re- 
mainder, which  can  be  but  once  ;  put  1  in  the  quotient, 
and  multiply  and  subtract,  and  the  next  remainder  is 
21373.  Then  point  o  IT'  the  4  in  the  divisor,  and  seek 
how  often  the  remaining  figures  may  be  had  in  21372, 
which  will  be  9  times  ;  put  9  in  the  quotient ;  multi- 
ply as  in  contracted  multiplication,  and  thus  proceed 
till  the  division  is  finished. 

I  have  set  down  the  w  ork  of  this  example  at  large, 
according  to  the  common  way,  that  thereby  the  learn- 
er may  see  the  reason  of  the  rule  ;  all  the  figures  on 
the  right -hand  side  of  the  perpendicular  line  being  whol- 
ly omitted. 

PRACTICAL    EXAMPLES. 

2.  Let  5171.59165  be  divided  hy  8.758615,  and  let 
it  he  required,  that  four  places  of  decimals  he  pointed 
oft'  in  the  quotient.  Jlns.  590.4577 

3.  Let  25.1367  be  divided  by  217.3543,  and  let 
there  be  five  places  of  decimals  in  the  quotient. 

Jins.  .11564. 
1.  Divide  7414.76717  by  2.756756,  and  let  there  be 
five  places  of  decimals  in  the  quotient. 

Ms.  2689.6flJS 


Extraction  of  the  Square  Root.  25 

5.  Divide  514.75498  by  12.34254,  and  let  there  be 
six  places  of  decimals  in  the  quotient. 

Jliis.  41.705757. 

6.  Divide  47194.379457  by  14.73495.  and  let  the 
quotient  contain  as  many  decimal  places  as  there  will 
be  integers,  or  whole  numbers,  in  it. 

JillS.  3202.88G9. 


CAA    CERVIX 

Extraction  of  the  Square  Root. 

If  a  square  number  be  given  ; 

J  O  find  the  Root  thereof,  that  is,  to  find  out  such  a 
number,  as  being  multiplied  into  itself,  the  product 
shall  be  equal  to  the  number  given ;  such  operation  is 
called,  The  Extraction  of  the  Square  Root ;  which  to 
do,  observe  the  following  directions. 

1st,  You  must  point  your  given  number;  that  is, 
make  a  point  over  the  unit's  place,  another  over  the 
hundred's,  and  so  over  every  second  figure  throughout. 

2dly.  Then  seek  tlie  greatest  square  number  in  the 
first  period  towards  the  left  hand,  placing  the  square 
number  under  that  point,  ami  the  root  thereof  in  the 
quotient,  and  subtract  the  said  square  number  from  the 
first  point,  and  to  the  remainder  bring  down  the  next 
point,  and  call  that  the  resolvend. 

3dly,  Then  double  the  quotient,  and  place  it  for  a  di- 
visor on  the  left  hand  of  the  resolvend ;  and  seek  how 
often  the  divisor  is  contained  in  the  resolvend  (re- 
serving always  the  unit's  place)  and  put  the  answer  in 
the  quotient,  and  also  on  the  right  hand  side  of  the 
divisor ;  then  multiply  by  the  figure  last  put  in  the 
quotient,  and  subtract  the  product  from  the  resolvend 
(as  in  common  division)  and  bring  down  the  next  point 

D 


Extraction  of  the  Square  Hoot. 

to  the  remainder  (if  there  he  any  more)  and  proceed 
as  before. 

Ji  Table  of  Squares  and  their  Hoots. 


.t    1 

1 

3 

8 

1    * 

■    1    •    j 

8    |     U 

|  Square  | 

1 

*    1 

» 

16 

23  |    36  | 

♦9  | 

fit  |  SI 

Example  1.  Let  4489  be  a  number  given)  and   let 
the  square  root  thereof  be  required. 


4489(67 
36 


127)889  Resolveml. 

S89  Product. 


First,  point  the  given  number,  as  before  directed^ 
then  by  the  little  table  foregoing,  seek  the  greatest 
square  number  in  44  (the  first  point  to  the  left-hand) 
which  you  will  find  to  be  36,  and  6  the  root ;  put  36 
under  44,  and  6  in  the  quotient,  and  subtract  36  from 
44,  and  there  remains  8.  Then  to  that  S  bring  down 
the  other  point  89,  placing  it  on  the  right-hand,  so  it 
makes  889  for  a  resolvond ;  then  double  the  quotient 
6,  and  it  makes  12  ;  which  place  on  the  left-hand  for  a 
divisor,  and  seek  how  often  12  in  88  (reserving  the 
unit's  plaee)  the  answer  is  7  times  ;  which  put  in  the 
quotient,  and  also  on  the  right-hand  side  of  the  divisor, 
and  multiply  127  by  7,  as  in  common  division,  and  the 
product  is  S89,  which  subtracted  from  the  resolvend, 
there  remains  nothing  ;  so  is  your  work  finished  ;  and 
the  square  root  of  4489  is  67  ;  which  root  if  you  multi- 
ply by  itself,  that  is  67  by  67,  the  product  will  be  4489, 
equal    to  the  given  square  number,  and  proves  the 


Extraction  of  the  Square  Root.  27 

work  to  be  right.     Had   there  been  any  remainder  it 
must  have  been  added  to  the  square  of  the  root  found. 

Example  2.  Let  106929  be  a  number  given,  and  M 
the  square  root  thereof  be  required. 

108929(327 


62)169         Resolvend, 
.  124         Product. 


647)4529     Resolvend. 
4529     ProdueU 


First,  point  your  given  number,  as  before  directed, 
putting  a  point  over  the  units,  hundreds,  and  tens  of 
thousands  ;  then  seek  what  is  the  greatest  square  num- 
ber in  10  (the  first  point)  which  by  the  little  iable 
you  will  find  to  be  9,  and  3  the  root  thereof;  put  9 
under  10,  and  3  in  the  quotient ;  then  subtract  9  out  of 
10,  and  there  remains  1 ;  to  which  bring  down  69,  the 
next  point,  and  it  makes  169  for  the  resolvend;  then 
double  the  quotient  3,  and  it  makes  6,  which  place  on 
the  left-hand  of  the  resolvend  for  a  divisor,  and  seek 
how  often  6  in  16 ;  the  answer  is  twice  :  put  2  in  the 
quotient,  and  also  on  the  right-hand  of  the  divisor 
making  it  62.  Then  multiply  62  by  the  2  you  put  in 
the  quotient,  and  the  product  is  124 ;  which  subtract 
from  the  resolvend,  and  there  remains  45 ;  to  which 
bring  down  29,  the  next  point,  and  it  makes  4529  for 
a  new  resolvend.  Then  double  the  quotient  32,  and  it 
makes  64,  which  place  on  the  left  side  of  the  resolv- 
end for  the  divisor,  and  seek  how  often  64  in  452, 
which  you  will  find  7  times  :  put  7  in  the  quotient,  and 
also  on  the  right-hand  of  the  divisor,  making  it  647, 
which  multiplied  by  the  7  in  the  quotient,  it  makes 


28  Extraction  of  the  Square  Boot. 

4529,  which  subtracted  from  the  resolrend,  there  re- 
mains nothing.  80  327  is  the  square  root  of  the  giv-. 
en  number. 

Note.  The  root  will  always  contain  just  so  many 
figures,  as  there  are  points  over  the  given  number  to 
J>e  extract!  (1 :  And  these  figures  will  he  whole  num- 
bers or  decimals  respectively?  according  as  the  points 
stand  over  whole  numbers  or  decimals. — The  method 
of  extracting  the  square  root  of  a  decimal  is  exactly 
the  same  as  in  the  foregoing  examples,  only  if  the 
number  of  decimals  be  odd,  annex  a  cipher  to  the  right 
hand  to  make  them  even,  before  you  begin  to  point. 
The  root  may  be  continued  to  any  number  of  figures 
you  please,  by  annexing  two  cyphers  at  a  time  to  each 
remainder,  for  a  new  resolvend. 

PRACTICAL    EXAMPLES. 

3.  It  is  required  to  extract  the  square  root  of 
2268741.  Jns.  1506.23.  Hem.  121871. 

t.  What  is  the  square  root  of  7596796  ? 

J$tl8.  2756.228.  Rem.  3212016.. 

5.  What  is  the  square  root  of  751427.5745  ? 

Jlns.  866.84.  Rem.  59889. 

6.  Extract  the  square  root  of  656714.37512. 

Jlns.  810.379.  Rem.  251479. 

7.  What  is  the  square  root  of  15241578750190521  ? 

Jlns.  123456789. 

8.  What  is  the  square  root  of  75.347  ? 

Jlns.  8.6802649729.     Rem.  24536226559. 

9.  What  is  the  square  root  of  .4325  ? 

Jlns.  .65764.     Rem.  96304. 

To  extract  the  Square  Root  of  a  Vulgar  Fraction. 

RULE. 

1.  Reduce  the  given  fraction  to  its  lowest  terms,  if 


Extraction  of  the  Square  Root. 


29 


it  be  not  in  its  lowest  terms  already ;  then  extract  the 
square  root  of  the  numerator  for  a  new  numerator, 
and  the  square  root  of  the  denominator  for  a  new  de- 
nominator. 

2.  If  the  fraction  will  not  extract  even,  reduce  it  to 
a  decimal,  and  then  extract  the  square  root. 

3.  When  the  number  to  be  extracted  is  a  mixed 
fraction,  reduce  the  fractional  part  to  a  decimal,  and 
annex  it  to  the  whole  number,  then  extract  the  square 
root. 

Example  1.  Extract  the  square  root  of  £££ 
First,  |$£  is  equal   to  •§§  in  its  lowest    terms,   the 

square  root  of  25  is  5,  and  the  square  root  of  36  is  6  5 

therefore  £  is  the  root  required. 

Example  2.  Let  seven-eighths  be  a  vulgar  fraction 
.  wliose  square  root  is  required. 


8)7.000 
61 


(.87500000(.935-l 

81 

183)650 
519 


1S65)10100 
9325 


1ST 01)77500 
71S16 


>6S1 


Reduce  this  |  to  a  decimal,  it  makes  .875 ;  (0  which 
annex  cyphers,  and  extract  the  square  root,  as  if  it 
was  a  whole  number.    So  the  root  is  .9351. 


d2 


30  Extraction  of  ike  Square  Root. 

Example  3.    Let  -^  be  a  vulgar  fraction,  whose 
square  root  is  required. 


)3.000000 

288 

(.00312500(.0559  Root. 

120 
90 

103)625 
525 

240 
192 

'     1109)10000 
9981 

480 
480 

19 

Example  4.  What  is  the  square  root  of  15  j  ? 
Here  |  reduced  to  a  decimal  is  .625,  which  annex 
ed  to  the  15  makes  15.625,  the  square  root  of  which 
is  3.95284.  Hem.  559344. 


PRACTICAL    EXAMPLES. 


FKACT1CAJL    tXAMrLfcb 

i.  What  is  the  square  root  of  |££  ? 
i.  What  is  the  square  root  of  ||£? 
'.  What  is  the  square  root  of  -pf? 

Ms.  .691S984,  &c. 
;.  What  is  the  square  root  of  29^\  ?  Ms.  5.4. 

I.  What  is  the  square  root  of  |? 

Ms.  3333.       Rem.  8W* 


Extraction  of  the  Cube  Root, 

CHAPTER  VIII. 

Extraction  of  the  Cube  Root. 

TO  extract  the  cube  root,  is  nothing  else  but  to  find 
such  a  number,  as  being  first  multiplied  into  itself, 
and  then  into  that  product,  produceth  the  given  num- 
ber; which  to  perform,  observe  the  following  direc- 
tions. 

is*,  You  must  point  your  given  number,  beginning 
with  the  unit's  place,  and  make  a  point,  or  dot,  over 
every  third  figure  towards  the  left-hand. 

2dbj,  Seek  the  greatest  cube  number  in  the  first 
point,  towards  the  left-hand,  putting  the  root  thereof 
in  the  quotient,  and  the  said  cube  number  under  the 
first  peiut,  and  subtract  it  therefrom,  and  to  the  re- 
mainder bring  down  the  next  point,  and  call  that  the 
resolvend. 

3f%,  Triple  the  quotient,  and  place  it  under  the  re- 
solvend; the  unit's  place  of  this  under  the  ten's  place 
of  the  resolvend ;  and  call  this  the  triple  quotient. 

tehly.  Square  the  quotient,  and  triple  the  square, 
and  place  it  under  the  triple  quotient;  the  units  of 
this  under  the  ten's  place  of  the  triple  quotient,  and 
call  this  the  triple  square. 

Bthly,  Add  these  two  together,  in  the  same  order  as 
they  stand,  and  the  sum  shall  be  the  divisor. 

Gthly,  Seek  how  often  the  divisor  is  contained  in 
the  resolvend,  rejecting  the  unit's  place  of  the  resol- 
vend (as  in  the  square  root,)  and  put  the  answer  in  the 
quotient 


Extraction  of  the  Cube  Rout. 

7'thly,  Cube  the  figure  lust  put  in  the  quotient,  and 
put  the  unit's  place  thereof  under  tht  unit's  plaee  of 
the  resolvend. 

Sthly,  Multiply  the  square  of  the  figure  last  put  in 
the  quotient  into  the  triple  quotient,  and  plaee  the 
product  under  the  last,  one  place  more  to  the  left- 
hand. 

Qthly,  Multiply  the  triple  square  by  the  figure  last 
put  in  the  quotient,  and  place  it  under  the  last,  one 
place  more  to  the  left-hand. 

\Othly,  Add  the  three  last  numbers  together,  in 
the  same  order  as  they  stand,  and  call  that  the  sub- 
trahend. 

Lastly,  Subtract  the  subtrahend  from  the  resolvend, 
and  if  there  be  another  point,  bring  it  down  to  the  re- 
mainder, and  call  that  a  new  resolvend,  and  proceed 
in  all  respects  as  before. 

Note.  To  square  a  number  is  to  multiply  that  num- 
ber by  itself.     And, 

To  cube  a  number  is  to  multiply  the  square  of  the 
number  by  the  number  itself. 

» 

Jl  Table  of  Cubes  and  their  Roots. 


Roots  1 

2 

3  |  4 

5 

6 

7 

8  |  9 

Cubes  1 

8 

27  |  64 

125 

216 

343 

512 j  729 

Extraction  of  the  Cube  Root.  3$ 

Example  1.  Let  314132  be  a  cubic  number,  whose 
loot  is  required. 


314132(68  Root. 
216 


98132  Resolvent! . 


18  Triple  quotient  of  6. 
108     Triple  square  of  the  quotient  0. 

1098  Divisor. 


512  Cube  of  8,  the  last  figure  of  the  root. 
1152     The  square  of  8,  by  the  triple  quotient. 
864      The  triple  square  of  the  quotient  6  by  8. 

98432  The  subtrahend. 


After  you  have  pointed  the  given  number,  seek 
what  is  the  greatest  cube  number  in  314,  the  first 
point,  which,  by  the  little  table  annexed  to  the  rule 
you  will  find  to  be  216,  which  is  the  nearest  that  is 
less  than  314,  and  its  root  is  6 ;  which  put  in  the  quo- 
tient, and  216  under  314,  and  subtract  it  therefrom, 
and  there  remains  98 ;  to  which  bring  down  the  next 
point,  432,  and  annex  it  to  98 ;  so  will  it  make  98432 
for  the  resolvend.  Then  triple  the  quotient  6,  it  makes 
18,  which  write  down  the  unit's  place,  8,  under  3,  the 
ten's  place  of  the  resolvend.  Then  square  the  quo- 
tient 6,  aud  triple  the  square,  and  it  makes  108,  which 
write  under  the  triple  quotient,  one  place  towards  the 
left-hand ;  then  add  those  two  numbers  together,  and 
they  make  1098  for  the  divisor.  Then  seek  how  often 
the  divisor  is  contained  in  the  resolvend,  (rejecting  the 
unit's  place  thereof)  that  is,  how  often  1098  iu  9843. 


8*  d  ruction  of  the  Cube  Hoot. 

which  is  S  times;  put  8  in  the  quotient,  and  the  cube 
(hereof  below  the  divisor,  the  unit's  place  under  the 
unit's  place  of  the  rcsolvend.  Then  square  the  8  last 
put  in  the  quotient,  and  multiply  64,  the  square  there- 
of*, by  the  triple  quotient  IS:  the  product  is  1102)  set 
this  under  the  cube  of  8,  the  units  of  this  under  the 
tens  of  that.  Then  multiply  the  triple  square  of  the 
quotient  by  8,  the  figure  last  put  up  in  the  quotient, 
the  product  is  864;  set  this  down  under  the  last  pro- 
duct, a  place  more  to  the  left-hand.  Then  draw  a 
line  under  these  three,  and  add  them  together,  and  the 
sum  is  08432,  which  is  called  the  subtrahend;  and  be- 
ing subtracted  from  the  rcsolvend,  the  remainder  is 
nothing;  which  shews  the  number  to  be  a  true  cubic 
number,  w  hose  root  is  6S ;  that  is,  if  68  be  cubed,  it 
will  make  314132. 

For  if  68  be  multiplied  by  68,  the  product  will  be 
4624;  and  this  product,  multiplied  again  by  68,  the 
last  product  is  314432,  which  shews  the  work  to  be 
right. 

Example  2.  Let  the  cube  root  of  5735339  be  re- 
quired. 

After  you  have  pointed  the  given  number,  seek  what 
is  the  greatest  cube  number  in  5,  the  first  point,  which, 
by  the  little  table,  you  w  ill  find  to  be  1 ;  which  place 
under  5,  and  1,  the  root  thereof,  in  the  quotient;  and 
subtract  1  from  5,  and  there  remains  4;  to  which  bring 
down  the  next  point,  it  makes  4735  for  the  rcsolvend. 
Then  triple  the  1,  and  it  makes  3;  and  the  square  of 
1  is  1,  and  the  triple  thereof  is  3;  which  set  one  under 
another,  in  their  or.der,  and  added,  makes  33  for  the 
divisor.  Seek  how  often  the  divisor  goes  in  the  resol- 
vend,  and  proceed  as  in  the  last  example. 


Extraction  of  the  Cube  Root,  35 


5735339(179  Root. 
1 


4735 


3  The  triple  of  the  quotient  1,  the  first  figure. 
3     The  triple  square  of  the  quotient  1. 


33  The  divisor. 


343  The  cube  of  7,  the  second  figure  of  the  root. 
147     The  square  of  7,  multipl.  in  the  triple  quot.  3. 
21       The  triple  square  of  the  quot.  multiplied  by  7. 

3913  The  subtrahend. 


822339  The  new  resolvend. 


51     The  triple  of  the  quot.  17,  the  two  first  fig. 
867       The  triple  square  of  the  quotient  17. 


8721  Divisor. 


729  The  cube  of  9,  the  last  figure  of  the  root. 
4131     The  squ.  of  9,  multipl.  by  the  triple  quo.  51. 
7S03       The  triple  square  of  the  quotient  867  by  9. 

822339  The  subtrahend. 


In  this  example,  33,  the  first  divisor,  seems  to  be 
contained  more  than  seven  times  in  473,  the  resolvend, 
after  the  unit's  place  has  been  rejected ;  but  if  you 
work  with  9,  or  8,  you  will  find  that  the  subtrahend 
will  be  greater  than  the  resolvend. 


30 


ruction  of  the  t 


Example  3.  Required  the  cube  ro 


22069310125(2805 

8 


1 fc069  Resolveud. 


G  Triple  of  3. 
\z    Triple  square  <>i'  2. 


126  Divisor. 

512  Cube  of  8. 
Square  of  s  by  f>. 

96  Triple  square  by  S. 


13952  Subtrahend. 


117810125  New  resolvcnd. 


8  l   Triple  of 

Triple  square  of  :s. 


23604  Divisor. 


8-10  Triple  of  280. 
m)  Triple  square  of  2S0. 


:<>  New  divisor. 


In  this  example 
2,  being 
tracted  from  (lie  re- 
solvend 14009.  the 
remainder  is  1 1 T  ; 
to  which  bring  down 
810,  the  3d.  point, 
and  it  makes  117810 
for  a  new  resolvend; 
and  the  next  divi- 
sor is  23604,  which 
you  cannot  have  in 
the  said  resolvend 
(the  unit's  place 
being  rejected:)  so 
you  must  put  0  in 
the  quotient,  and 
seek  a  new  divi- 
sor (after  you  have 
brought  down  your 
last  point  to  the  re- 
solvend;) which  new 
divisor  is  2852640; 
and  you  will  find  it 
to  be  contained  ."> 
times.  So  proceed  to 
finish  the  rest  of  the 
work. 


l  M  Tube  of  five. 
21000     Square  of  5  by  840. 
l  IT nooo      Triple  square  by  5. 


£17810125  Subtrahend. 


Extraction  of  the  Cube  Root. 

Note.  The  root  will  always  contain  just  so  many 
figures,  as  there  are  points  over  the  given  number  to 
be  extracted;  and  these  figures  will  be  whole  numbers 
or  decimals  respectiv.lv,  according  as  the  points  stand 
over  whole  numbers  or  decimals.  The  method  of 
extracting  the  cube  root  of  a  mixed  number,  or  deci- 
mal, is  the  same  as  in  the  above  examples;  only  the 
number  of  decimals  must  be  made  to  consist  of  three, 
six  or  nine,  &c.  figures,  by  annexing  cyphers. 

PRACTICAL    EXAMPLES. 

Example  4.  What  is  the  cube  root  of  32461759  ? 

Ms.  31 9 

5.  What  is  the  cube  root  of  84604519  ?        Ms.  439 

6.  What  is  the  cube  root  of  259697989  ?      Ms.  638 

7.  What  is  the  Cube  root  of  '33917056  ? 

Ms.  295.9.  Rem.  8995931 

8.  What  is  the  cube  root  of  93759.57507  ? 

Ms.  45.42.  Rem.  59186982 

9.  Required  the  cube  root  of  .401719179. 

Ms.  .737.  Rem.  1403626 

10.  Required  the  cube  root  of  .0001416  ? 

Ms.  .052.  Rem.  992 

11.  Required  the  cube  root  of  122615327232. 

Ms.  496  S 

12.  What  is  the  cube  root  of  705.919947284? 

Ms.  8.904.  Rem.  20. 

13.  What  is  the  cube  root  of  }  «„   Ai~~<.^r- 

1531328.215978518623  ?   \  MS'  li5"  «25 

14.  The  cube  root  of  .57345  is  required. 
Ms.  .S308.  Rem.  8045888 


To  extract  the  Cube  Root  of  a  Vulgar  Fraction. 

rule. 

1.  Reduce  the  given  fraction  to  its  lowest  terms,  if 
it  be  u.o%  in  it*  lowest  terms  alreadv ;  then  extract 

E 


3$  Extraction  of  the  Cuhe  Hmt. 

the  cube  root  of  the  numerator  tor  a  new  numerator, 
ami  the  cube  root  of  the  denominator  for  a  new  deno- 
minator. 

2.  If  the  fraetion  will  not  extract  even,  reduce  it  to 
a  decimal,  and  then  extract  the  cube  root. 

3.  When  the  number  to  be  extracted  is  a  mixed 
fraction,  reduce  the  fractional  part  to  a  decimal,  and 
annex  this  decimal  to  the  whole  number,  then  extract 
the  cube  root. 

Example  1.  What  is  the  cube  root  of  |^|  ? 
First.  ^*|  is   equal  to   f  £  in  its  lowest  terms,  the 
cube  root  of  27  is  3,   and  the   cube  root  of  64<  is  4; 
therefore  the  cube  root  of  -||  is  J,  the  answer. 

Example  2.  Let  ?|y  be  a  vulgar  fraction,  whose 
cube  root  is  required. 

By  the  first  rule  of  Chapter  II.  reduce  the  vulgar 
fraction  to  a  decimal. 

276)5.000000000(.018U5912 
276 


22  tO 


1640 


2600 


1160 

060 


Extraction  of  the  Cube  Moot.  .39 


.018115942(.262  Root. 

8 


10 115  Resolvend. 


6  Triple  of  2. 
12     Triple  square  of  2. 


126  Divisor. 


216  Cube  of  6. 
216     Square  of  6  by  the  triple  of  ,2. 
72     Triple  square  by  6. 


9576  Subtrahend. 


539942  Resolvend. 


78  Triple  of  26. 
2028     Triple  square  of  26, 


20358  Divisor. 


S  Cube  of  2. 
312     Square  of  2  by  78. 
j.056      Triple  square  2028  by  2. 

408728  Subtrahend. 


131211  Remainder. 

You  may  prove  the  truth  of  the  work,  by  cubing  the 
root  found,  as  was  shewn  in  the  first  example:  and  if 
any  thing  remains,  add  it  to  the  said  cube,  and  the 
sum  will  be  the  given  number,  if  the  work  is  rightly 
performed. 


Multiplication  of  Feet,  %'c. 

Example  3.  What  is  the  cube  root  of  56623  TVT-; 
Here  ^  reduced  to  a  decimal  is  .104,  which  annex- 
ed to  36623  makes  36623.104,  the  cube  root  of  which 
8.4. 

PRACTICAL    EXAMPLES. 

4.  What  is  the  cube  root  of  i£56  ?  Ms.  7163S 
Bf.  Required  the  cube  root  of  j£te.             Ms. 

6.  What  is  the  cube  root  of  -fifor  Ms.  .}. 

7.  What  is  the  cube  root  of  |  ?  Ms.  .82207 

5.  What  is  the  cube  root  of  |J  ?  Ms.  .98591 
9.  What  is  the  cube  root  of  5333|?  Jins.  .17.471 

10.  What  is  the  cube  root  of  108220l|? 

Ms.  loi.or 


CHAPTER  IX. 

Multiplication  of  Fee*,  Inches,  and  Part's;  ov 
Duodecimals, 

THE  multiplication  of  feet  and  inches  is  generally 
called  duodecimals,  because  ever}7  superior  place  is  i2 
limes  its  liext  inferior  in  this  scale  of  notation.  This 
way  of  conceiving  an  unit  to  be  divided,  is  chiefly  in 
use  among  artificers,  who  generally  take  the  linear  di- 
mensions of  their  work  in  feet  and  inches :  It  is  like- 
wise called  cross  midtiplication,  because  the  factors 
are  sometimes  multiplied  crosswise. 

RULE    I. 

1.  Under  the  multiplicand  write  the  corresponding 
denominations  of  the  multiplier;  that  is,  feet  under 
feet,  inches  under  inches,  parts  under  parts,  &c. 


Multiplication  of  Feetf  $c.  41 

2.  Multiply  each  term  in  the  multiplicand,  begin- 
ning at  the  lowest,  by  the  feet  in  the  multiplier ;  write 
each  result  under  its  respective  term,  observing  to 
carry  an  unit  for  every  12,  from  each  lower  denomi- 
nation to  its  next  superior. 

3.  In  the  same  manner  multiply  every  term  in  the 
multiplicand  by  the  inches  in  the  multiplier,  and  set 
the  result  of  each  term  one  place  removed  to  the  right- 
hand  of  those  in  the  multiplicand. 

4.  AVork  in  a  similar  manner  with  the  parts  in  the 
multiplier,  setting  the  result  of  each  term  removed 
two  places  to  the  right-hand  of  those  in  the  multipli- 
cand. Proceed  in  like  manner  with  the  rest  of  the 
denominations,  and  their  sum  will  be  the  answer  re- 
quired. 

Examfle  1.  Let  7  feet  9  inches  be  multiplied  by  3 
feet  6  inches. 

F.      I. 

Multiplicand    7   ••    9 

'    Multiplier        3    -6 

23-3       Fts. 
3    ••  10   ••    6 


Product  27    -.     1    ..    6 


First,  Multiply  9  inehes  by  3,  saying,  3  times  9  is 
27  inches,  which  make  2  feet  3  inches ;  set  down  3 
under  inches,  and  carry  2  to  the  feet,  saying,  3  times 
7  is  21,  and  2  that  I  carry  make  23;  set  down  23  un- 
der the  feet. 

Then  begin  with  6  inches,  saying,  6  times  9  is  54 
parts,  which  is  4  inches  and  6  parts ;  set  down  6  parts, 
and  carry  4,  saying,  6  times  7  is  42,  and  4  that  I  carry- 
is  46  inches,  which  is  3  feet  10  inches;  which  set 
down,  and  add  all  up  together,  and  the  product  is 
27  feet  1  inch  6  parts. 

E    2 


Multiplication  of  Feet.  • 

Example  2.    Let  r  feet  0  inches  9  parts  be  multi- 
plied b)  3  leet  5  inches  3  pari-. 


F. 

Multiplicand  7     .. 
Multiplier       3     .. 

I. 

5 
5 

P. 

..      9 

> 

3     .. 

5 
1 

1 

..3        V 

..    9        T. 

..  10  ..  5  ..  .: 

Product  26     .. 

8 

..     0  ..  2  ..  3 

In  this  example,  I  first  begin  with  3  feet,  and  there- 
by multiply  7  feet  5  inches  and  9  parts  :  First,  I  say, 
3  times  9  is  27  parts,  that  is  2  inches  and  3  parts  5 
set  down  3  under  the  parts,  and  carry  2,  saying,  3 
times  5  is  15,  and  2  I  carry  is  17,  that  is,  1  foot  5 
inches ;  set  down  5  inches,  and  carry  1,  and  say,  3 
times  7  is  21,  and  1, 1  carry  is  22;  set  down  22  feet: 
Then  begin  with  5  inches,  saying,  5  times  9  is  45, 
which  is  45  seconds,  which  makes  3  parts  and  9  se- 
conds ;  set  down  9  seconds  a  place  towards  the  right- 
hand,  and  carry  3  parts,  saying,  5  times  5  is  25,  and 
3  I  carry  is  28,  which  is  2  inches  and  4  parts ;  set 
•down  4  parts  and  carry  2,  saymg  5  times  7  is  35,  and 
2,  I  carry  is  37,  whieh  h  3  feet  1  inch;  set  down  3 
feet  1  inch  ;  and  begin  to  multiply  by  3  parts,  say- 
ing, 3  times  9  is  27  thirds,  that  is,  2  seconds  and  3 
thirds  ;  set  down  3  thirds,  and  carry  2,  saying  3  times 
5  is  15,  and  2, 1  carry  is  17,  that  is,  1  part  and  5  se- 
conds -,  set  down  5  seconds,  and  carry  1,  saying  3  times 
7  is  21,  and  1, 1  carry  is  22,  whieh  is  1  inch  and  40 
parts,  which  set  down,  and  add  all  up,  and  the  prx>~ 
duct  is  25  feet  8  jnehes  6  parts  2  seconds  3  thirds. 


When  h 


•Multiplication  of  Feet,  $c. 


43 


RULE    II. 


'ten  the  Feet  in  the  Multiplicand  are  expressed  by 
a  large  number. 

Multiply  first  by  the  feet  in  the  multiplier,  as  be- 
fore. Then,  instead  of  multiplying  by  the  inches  and 
parts,  &c.  proceed  as  in  the  Rule  of  Practice,  by  tak- 
ing such  aliquot  parts  of  the  multiplicand  as  corres- 
pond with  the  inches  and  parts,  &e.  of  the  multiplier. 
Then  the  sum  of  them  all  will  be  the  product  required. 

Example  3.  Let  75  feet  7  inches  be  multiplied  by 
9  feet  8  inches. 


In. 

F. 

I. 

$ 

75 

..  7 

Multiplicand 

J 

9 

..  8 

Multiplier. 

680 

..  3 

25 

..  2. 

..  4 

25 

..  2 

..  4 

I 

'roduet 

730  .. 

7  . 

.   S 

Multiply  by  9  feet,  first,  as  above  directed  ;  ihea 
instead  of  multiplying  by  the  8  inches,  let  them  be  di- 
vided into  aliquot  parts  of  a  foot,  as  4  and  4,  because 
4  is  the  third  part  of  12.  So,  if  you  take  the  third 
part  of  75  feet  7  inches,  and  set  it  down  twice,  and 
add  all  together,  the  sum  will  be  730  feet  7  inches  8 
parts.  To  take  the  third  part,  say,  how  often  3  in  7, 
which  is  twice  ;  set  down  2  ;  then  because  twice  3  is 
6,  say,  6  out  of  7,  and  there  remains  1,  for  which  you 
laiust  add  10  to  the  5,  and  it  makes  13  ;  then  the  threes 
in  15  are  5  times:  set  down  5:  and,  because  three 
times  5  is  15,  there  is  0  remains.  Then  go  to  the  7 
inches,  saying,  the  three  in  7  are  twice  :  set  down  '2 


41 


Multiplication  of  Fee 


in  the  inches ;  and  because  <\\  lake  fi  out  of  7, 

arid  there  remains  l  inch,  which  is  12  parte;  then  threes 

in  1  Jure  Mimes,  and  0  remains.  So  the  third  part  of 
75  feet  7  inches  is  $0  feel  2  incites  4  parts;  which  set 
twice  over,  and  add  them  together  as  in  the  example. 

Example  l.     Let  37  feet  7  inches  5  parts  he  multi- 
plied by  4  feet  S  inches  6  parts. 


I.      P. 


I    ..    0) 

1  ••  05 


I.    p. 

0    ..    6 


F.      I. 

P. 

37    ..    7 

-    13 

Multiplicand 

4-8 

-    6 

Multiplier. 

150    -    5 

-    8 

S. 

12    ••    6 

-    5 

-    8 

12    »    6 

-    5 

-    8          T. 

1-6 

-    9 

..8-6 

Product       177 


In  this  example  I  first  multiply  by  4  feet  as  usual.— 
Then  for  the  8  inches  I  say  4  inches  is  the  third  of  a 
foot,  therefore  I  take  the  third  part  of  37  feet  7  inches 

5  parts,  which  is  12  feet  6  inches  5  parts  8  seconds,  and 
set  it  down  twice.  Then  for  6  parts,  I  say,  6  parts 
are  the  eighth  of  4  inches,  because  12  parts  make  1 
inch  ;  hence  it  follows,  that  whatever  be  the  value,  or 
product,  by  4  inches,  the  value  of  6  parts  will  be  one- 
eighth  thereof;  therefore  I  take  one-eighth  of  12  feet 

6  inches  5  parts  8  seconds,  and  find  it  to  be  1  foot  6 
inches  9  parts  8  seconds  6  thirds  ;  so  that  the  sum  of 
the  whole  is  177  feet  1  inch  5  parts  6  thirds. 


Multiplication  of  Feet,  $c* 


46* 


RULE    III. 


When  the  Feet  both  in  the  Multiplicand  and  Multiplier 
are  large  numbers. 

Multiply  the  feet  only  into  each  other:  Then,  for 
the  inches  and  parts  in  the  multiplier,  take  parts  of 
the  feet,  inches,  &c.  of  the  multiplicand  :  And,  for  the 
inches  and  parts  of  the  multiplicand,  take  parts  of  the 
feet  o?Uy  in  the  multiplier.  The  sum  of  all  will  be  the 
product. 


Example  5.    Let  75  feet  9  inches  be  multiplied  by 
17  feet  7  inches. 


I. 

p. 

I. 

0 

i 

T 

75 
17 

.»    9 
••   7 

525 

75. 

P. 

i 

1 

37 

••    10 

••    6 

6 

-.      3 

••    9 

12 

..      9 

•• 

•od 

net 

1331 

..    11 

..    3 

I. 

F. 

6 

l 

17    • 

T 

3 

l 

8    • 

•    6 

4    . 

•    3 

12 


In  this  example,  because  there  are  more  than  12 
feet  in  the  multiplier,  I  first  multiply  the  75  feet  by  17 
feet.  Then,  I  say,  6  inches  are  the  half  of  a  foot,  and 
take  the  half  of  75  feet  9  inches,  which  is  37  feet  10 
inches  6  parts  ;  but  I  ought  to  take  the  parts  for  7 
inches,  therefore  I  say  1  inch  is  the  sixth  of  6  inches, 
and  take  the  sixth  part  of  37  feet  10  inches  6  parts, 
which  I  find  to  be  6  feet  3  inches  9  parts.  Then,  be- 
cause there  are  9  inches  in  the  multiplicand,  I  take 
parts  with  them  out  of  the  17  feet  in  the  multipli- 
er, saying,  6  inches  are   the  half  of  a  foot,  I  there- 


Multiplication  of  7< 


ke  the  half  of  17  feet*  which  is  8  fee$  6  inches; 

inches  left,  and  whatever  the 

:  by    6  inches  may    be,   that  by  3  inches   must 

half  thereof  \  I  inches  are  the  lialt;  and 

the  half  of  8  feet  6  inches,  which  is  4  feci  3  in- 

li<^.  the   sum   of  these  is  12  feet  9  inches,  which   1 

place  under  the  former   parts,  and    the  sum   of  the 

whoii  feet  11  inches  3  parts. 

.  6.  Let  3JI  feet  l  inches  7  parts   be  mul 
tiplied  by  36  feet  7  inches  5  parts. 

F. 

36 


1..G 
0..3 


13-.  9 


In.  P. 

6  •• 

i 
I 

1-.  0 

0.-4, 

CI 

~5 

l 
T 

i 
T 

F.    I.    P. 

311  ..4-7 

36-" 

1866 

933 .        I.     P. 

8. 

150-.    8  ••    \  •• 

6 

25   .11  ••  4  •• 

7 

T 

8-    7  -  9  •• 

6 

.4 

2»    1..11  .. 

4  • 

.7 

13-    9 

I.  P. 

4..0 

T 

0-6 

0-1 

i 

Product   11402  ••    2  «  4  ••  11  ••  11 


In  this  example  I  first  multiply  the  feet  as  in  •  - 
pic  5th.     Then  I  say  6  inches   are   the  half  of  a  foot, 
and  take  the  half  of  311  feet  4  inches  7  parts,  which 

1  find  to  be  155  feet  8  inches  3  parts  6  seconds;  then, 
as  1  inch  is  one-sixth  of  6  inches.  I  therefore  take  pne 
sixth  of  155  Feet  8  inches  3  parts  6  seconds,  which 
is  25  i'eet  1  J.  inches  4  pai  ads :  Then,  because  4 
parts  are  one  third  of  an  inch,  I  take  one  third  of  25 

I  inches  4  parts  7  seconds,  and  find  it  to  be  8  feet 

7  inches  9  parts  6  seconds  4  thirds ;  and  as  I  have  one 

ft,  I  say  1  is  the  fourth  of  4,  and  take  the  fourth 

of  s  feet  7  inches  9  parts  6  Seconds  i  thirds,  which  is 

2  feet  1  inch  11  parts  4  seconds  7   thirds. — Then  for 


Multiplication  of  Feat,  $c.  47 

the  four  inches  in  the  multiplicand,  I  take  a  third  part 
of  36,  the  feet  in  the  multiplier  ;  because  4  inches  are 
one-third  of  a  foot ;  and  6  parts  are  the  eighth  of  4 
inehes,  1  therefore  take  the  eighth  of  12  feet  which  is 

I  foot  6  inches ;  then  I  have  1  part  left,  which  is  the 
sixth  of  6  parts,  so  I  take  the  sixth  of  1  foot  6  inches, 
which  is  3  inches.  The  sum  of  these  parts  is  13  feet 
9  inches,  which  I  place  under  the  former  parts,  and 
add  them  together,  so  that  the  whole  is  11402  feet  2 
inches  4  parts  11  seconds  11  thirds. 

PRACTICAL    EXAMPLES. 

7.  Let  97  feet  8  inches  he  multiplied  hy  8  feet  9 
inches.  Jlns.  854  feet  7  inches. 

8.  Let  87  feet  5  inehes  he  multiplied  by  35  feet  8 
inches.  Jlns.  3117  feet  10  inches  4  parts. 

9.  Let  259  feet  2  inches  be    multiplied  by  48  feet 

II  inches.  rfns.  12677  feet  6  inches  10  parts, 
10  Multiply  179  feet  3  inches  by  38  feet  10  inehes. 

•ins.  6960  feet  10  inches  6  parts. 

11.  Multiply  246  feet  7  inches  by  46  feet  4  inches. 

Jlns.  11425  feet  0  inches  4  parts. 

12.  Multiply  246  feet  7  inches  by  36  feet  9  inches. 

Jlns.  9061  feet  11  inches  3  parts. 

13.  Multiply  257  feet  9  inches  by  3W  feet  11  inch- 
es. Jlns.  10288  feet  6  inches  3  parts. 

14.  Let  8  feet  4  inches  3  parts  5  seconds  6  thirds 
he  multiplied  by  3  feet  3  inches  7  parts  8  seconds  2 
thirds.  Jins.  27  feet  7  inches  3  parts  5  seconds 

1  third  8  fourths  8  fifths  11  sixths. 

15.  Multiply  321  feet  7  inehes  3  parts  by  9  feet  3 
inches  6  parts.  Jlns.  2&SS  feet  2  inches  10  pads 

4  seconds  6  thirds, 

16     Multiply  42  feet  7   inches  8  parts   by  7  feet  3 

inches  6  parts.        Jlns.  310  feet  10  inches   10   parts 

10  second?* 


Of  C  Scale,  $c. 

Mult iply    l„M   fool  7  inches  9  parts  by  14  feet 
fi  iiielit Ji  3  pa 

I  feet  l  inch  1  part  0  seconds  6  thirds. 
Muhiph  to   inches  8   parts  by  18  feet 

•  indie*  i  pert*. 

99  i\  el   0  inches  0  parts  10  seconds  8  thirds. 
19.  Multiply  207  feet  7  inches  10  parts  by  23  feet 
9  inches  7  parts.  - 

Jus.  60(1.3  fefet  0  inches  5  parts  0  seconds  10  thirds. 
20  Multiply  317  feet  0  inches  7  parts  by  37    I 
inches  |  parts. 

Jn<.  liuio  feet  9  inches  11  parts  1  second  3  thirds. 


CHAPTER  X. 


nation  of  the  line  of  numbers  on  Gusrsn^s 
Scale,  and  tlte  construction  and  use  of  the  Common 
Diagonal  Scale. 


THE  line  of  numbers  on  the  two  feet  Gunter's  Scale, 
marked  Number,  is  numbered  from  th»  left-hand  of 
the  Scale  towards  the  right,  with  the  figures  1,  2,  3, 
4,  5,  6,  7,  8,  9 ;  1,  which  stands  exactly  in  the  middle 
of  the  Scale  :  the  numbers  then  go  on  2,  3,  4,  5,  6,  7, 
8,  9 ;  10,  which  stands  at  the  right-hand  end  of  the 
Scale. 

These  two  equal  parts  of  the  Scale  are  also  equally 
divided,  the  distance  between  the  first,  or  left-hand  1, 
and  the  first  2,  3,  4.  &,e.  is  exactly  equal  to  the  dis- 
tance between  the  middle  I,  and  the  numbers  2,  3,  4, 
&.C  which  follow  it 


Of  Gunter'>s  Scale,  $c.  4S 

The  subdivisons  of  these  two  equal  parts  of  the 
scale  are  likewise  similar,  viz.  they  are  each  one 
tenth  of  the  primary  divisions,  and  are  distinguished 
by  lines  of  about  half  the  length  of  the  primary  di- 
visions. These  subdivisions  are  again  divided  into 
ten  parts,  where  room  will  admit,  and  where  that  is 
not  the  case,  the  units  must  be  estimated,  or  guessed 
at  by  the  eye,  which  is  easily  done  by  a  little  practice. 

The  primary  divisions  on  the  second  part  of  the 
scale  are  estimated  according  to  the  value  set  upon 
the  unit  on  the  left-hand  of  the  scale :  Thus,  if  you 
call  the  unit  on  the  left-hand  of  the  scale  1,  then  the 
first  1,  2,  3,  4,  &c.  stand  for  1,  2,  3,  4,  &c.  the  middle 
1  is  10;  and  the  2,  3,  4,  &c.  following,  stand  for  20, 
30,  40,  &c.  and  the  10  at  the  right-hand  is  100. — If 
you  call  the  unit  on  the  left-hand  of  the  scale  10, 
then  the  first  1,2, 3,  4,  &c.  stand  for  10,  20,  30,  40,  &c„ 
the  middle  1  will  be  100 ;  and  the  2,  3,  4,  &e.  follow- 
ing will  be  200,  300,  400,  &c.  and  the  10  at  the  right- 
hand  will  be  1000 — If  you  call  the  unit  on  the  left- 
hand  of  the  scale,  100,  then  the  first  1,  2,  3, 4,  &e. 
will  stand  for  100,  200,  300,  400,  &c.  the  middle  1  will 
be  1000 ;  and  the  2,  3,  4,  &e.  following,  2000,  3000, 
4000,  &c.  and  the  10  at  the  right-hand  end  will  be 
10000. — Lastly,  if  you  consider  the  unit  on  the  left- 
hand  of  the  scale  as  one-tenth  of  an  unit,  then  the 
first  1,  2,  3,  4,  &c.  will  be  ^,  ^  T%,  T4^,  &e.  the  mid- 
dle 1  will  stand  for  an  unit,  and  the  2,  3,  4,  &c.  fol- 
lowing it  will  be  2,  3,  4,  &c.  and  the  10  at  the  right- 
hand  end  of  the  scale  will  stand  for  10. 

From  the  above  description  it  will  be  easy  to  find 
the  divisions  representing  any  given  number.  Sup- 
pose 12  was  required  ;  take  the  division  at  the  figure 
1 ,  in  the  middle  of  the  scale,  for  the  first  figure  of  12 ; 
then  for  the  second  figure,  count  two  of  the  longer 
strokes  to  the  right-hand,  and  this  last  is  the  point 
representing  12,  where  there  is  a  brass  pin. — If  34 
were  required :  Call  the  figure  3  on  the  right -hand 


/ 


of  a 


.   and  count    forward    four  of  the 

rUions  towards  the  right-hand:  if  340   were 

it    must   be  found    in  the   same  manner,     if 

the  point  i  ng  345  were  required,  lind  3  to  as 

then  the  middle  distance   between  the   point  of 

«h>.  and  the  point  representing  330,  will  be  the  point 

1>\  the  line  of  numbers  and  a  pair  of  compasses  al- 
'I  the  problems  in   mensuration  may  be  readily 

for  they  in  general  depend   upon    proportion . 

Ami,  as  in  natural  numbers,  the  quotient  of  the  first 
term,  of  any  abstract  proportion,  by  the  second,  is 
equal  to  the  quotient  of  the  third  term  by  the  fourth  ; 
so  in  logarithms  (for  the  line  of  numbers  is  a  loga- 
rithmical  line)  the  difference  between  the  first  and  se- 
cond term,  is  equal  to  the  difference  between  the  third 
and  fourth  :  consequently,  on  the  line  of  numbers,  the 
distance  between  the  first  and  second  term,  will  be  c- 
qual  to  the  distance  between  the  third  and  fourth. — 
And  for  a  similar  reason,  because  four  proportional 
quantities  are  alternately  proportional,  the  distance 
between  the  first  and  third  term  will  be  equal  to  the 
distance  between  the  second  and  fourth.  Hence  the 
following 

GENERAL    RULE. 

The  extent  of  the  compasses  from  the  first  term  to 
the  second,  will  reach,  in  the  same  direction,  from  the 
third  to  the  fourth  :  Or,  the  extent  of  the  compasses 
from  the  first  term  to  the  third,  will  reach  in  the  same 
direction,  from  the  second  to  the  fourth. 

By  the  same  direction  must  be  understood,  that  if 
the  second  term  lie  on  the  right-hand  of  the  first,  the 
fourth  term  will  lie  on  the  right-hand  of  the  third,  and 
the  contrary.     Hence, 

1.  To  find  the  product  of  two  Niiinbers. 
As  an  unit  is  to  the  multiplier,    so  is    the  multipli- 
cand to  the  product. 


Of  the  Diagonal  Scale. 


51 


II.  To  divide  one  Number  by  another. 
As  the  divisor  is  to  the  dividend,  so  is  an  unit  to  the 
quotient. 

III.  To  find  a  mean  proportional  between  two  Numbers. 
Because  the  distance  between  the  first  and  second 
term,  is  equal  to  the  distance  between  the  third  and 
fourth  ;  therefore,  if  you  divide  the  space  between  the 
point  representing  the  first  term,  and  that  represent- 
ing the  fourth,  into  two  equal  parts  :  the  middle  point 
must  necessarily  give  the   mean  proportional  sought. 

IV.  To  extract  the  Square  Root. 

The  square  root  of  a  quantity  is  nothing  more  than 
a  mean  proportional  between  an  unit  aud  the  given 
number  to  be  extracted  ;  the  unit  being  the  first  term, 
and  the  number  to  be  extracted  the  fourth ;  therefore 
it  may  be  done  by  the  preceding  direction. 

Note.  These  rules  are  all  applied  in  the  succeed- 
ing parts  of  the  book. 


Of  the  Diagonal  Scale. 
The  diagonal  scale,  usually  placed  on  the  two  feet 
Gunter's  scale,  is  thus  contracted. 
A.  B 


Draw  eleven  lines  of  equal  length  and  at  equal  dis- 
tances from  each  other,  as  in  the  above  figure;  divide 
the  two  outer  lines,  as  AD,  CE,  into  any  conveni- 
ent number  of  equal  parts,  according  to  the  largeness 
you  intend  your  scale.  Join  these  parts  by  straight 
lines,  as  AC,  Bo,  &c.  first  taking  care  that  the  cor- 
ners C,  A.  D,  E,  are  ail  square.    Again  divide  the 


Of  the  Carpenter's  Rule. 

lengths  AB,  and  Co,  into  to  equal  parts,  join  these 
riz.  from  the  point  A  to  the 
first  division  in  Co,  ami  from  the  first  division  in  AH, 
fo  the  second  division  in  Co,  &c.  as  in  the  figure,  and 
number  the  several  divisions. 

The  chief  use  of  sue!:  a  scale  as  this,  is  to  lay  down 

any  line  from  a  given  measure;  or  to    measure   any 

and   thereby  to   compare  it  with  others.     If  the 

divisions  in  oE  he  ealled  units,  the  small  divi- 
sions in  v_'o  will  be  LOths,  and  the  divisions  in  the  al- 
titude oB  will  be  100th  parts  of  an  unit.     If  the  large 

us  be  tens,  the  others  will  be  units,  and  tenth 
parts.  If  the  large  divisions  be  hundreds,  the  others 
will  be  tens  and  units,  &.c.  each  set  of  divisions  be- 
ing tenth  parts  of  the  former  ones. 

For  example,  suppose  it  were  required  to  take  off 
244  from  the  scale :  fix  one  foot  of  the  compasses  at 
.2  of  the  larger  divisions  in  oE,  and  extend  the  other 
to  the  number  4  in  Co  3  then  move  both  points  of  the 
compasses  by  a  parallel  motion,  till  you  come  at  the 
fourth  long  line,  taking  care  to  keep  the  right-hand 
point  in  the  line  marked  2  :  then  open  the  compasses 
a  small  matter,  till  the  left-hand  foot  reaches  to  the 
intersection  of  the  two  lines  marked  1,  4,  and  you  have 
the  extent  of  the  number  required.  In  a  similar  man- 
ner any  other  number  may  be  taken  oft*. 


CHAPTER  XI. 

Description  and  use  of  the  common  Carpenter's 
Rule. 
TlllS  rule  is  generally  used  in  measuring   of  timber, 
and  artificers  works  :  and  is  not  only  useful  in   taking 
dimensions,  but  in   casting  up  the  contents  of  such 
work. 

It  consists  of  two  equal  pieces  of  box,  each  one  foot 
in  length,  connected  together  by  a  folding  joint;  in  one 
of  the^e  equal  pieces  there  is  a  slider,  and  four  lines 


Of  the  Carpenters  Rule,  5% 

marked  at  (he  right-hand  A,  B,  C,  D  ;  two  of  these 
lines,  B,  C,  are  upon  the  slider,  and  the  other  two,  A, 
J},  upon  the  rule.  Three  of  these  lines,  viz.  A,  B,  C, 
are  called  double  lines,  because  they  proceed  from  1  to 
10  twice  over;  these  three  lines  are  all  exactly  alike, 
both  in  numbers  and  division.  They  are  numbered 
from  the  left-hand  towards  the  right  1,  2,  3,  4,  5,  6,  7, 
8,  9,  1  which  stands  in  the  middle ;  the  numbers  then 
go  on,  2,  3,  4,  5,  6,  7,  8,  9,  10  which  stands  at  the 
right-hand  end  of  the  rule.  These  numbers  have  no 
determinate  value  of  their  own,  but  depend  upon  the 
value  you  set  on  the  unit  at  the  left-hand  of  this  part 
of  the  rule ;  thus  if  you  call  it  1,  the  1  in  the  middle 
will  be  10,  the  other  figures  which  follow  will  be  20, 
30,  &e.  and  the  10  at  the  right-hand  end  will  be  100. 
If  you  call  the  first,  or  left-hand  unit  10,  the  middle  1 
will  be  100,  and  the  following  figures  will  be  200,  300, 
400,  &c.  and  the  10  at  the  right-hand  end  will  be  1000. 
Or,  if  you  call  the  first,  or  left-hand  unit,  100,  the 
middle  1  will  be  1000,  and  the  following  figures  2000, 
3000,  4000,  &c.  and  the  10  at  the  right-hand  10,000. 
Lastly,  according  as  you  alter,  or  number,  the  large  di- 
visions, so  you  must  alter  the  small  divisions  propor- 
tionality. 

The  fourth  lineD,  is  a  single  line,  proceeding  from 
4  to  40 :  it  is  also  called  the  girt-li?ie,  from  its  use  in 
easting  up  the  contents  of  trees  and  timber.  Upon  it 
are  marked  WG  at  17.15,  and  AG  at  18.95,  the  wine 
and  ale  guage  points,  to  make  it  serve  the  purpose  of 
a  guaging-rule. 

The  use  of  the  double  lines  A  and  B,  is  for  work- 
ing the  rule  of  proportion,  and  finding  the  areas  of 
plane  figures.  And  the  use  of  the  girt-line  D,  and 
the  other  double  line  C,  is  for  measuring  of  timber. — 
On  the  other  part  of  this  side  of  the  rule,  there  is  a 
table  of  the  value  of  a  load,  or  50  cubic  feet,  of  tim- 
ber, at  all  prices,  from  6  pence  to  24  pence,  or  two 
shillings,  per  foot. 

f  2 


Of  the  Carpenters  Ride. 

On  tike  other  side  of  the  rule  arc  several  plane 
scales  divided  into  12th  parts,  marked  inch,  |,  £,  \, 
gmtying,  that  the  inch,  J  inch,  &c.  are  each  di- 
vided into  13  parts.  These  scales  are  useful  for  plan- 
ning dimensions  that  are  taken  in  feet  and  inches. 
The  edge  of  the  ruler  is  divided  into  inches,  and  each 
of  these  inches  into  eight  parts,  representing  half  in- 
ches, quarter  inches,  and  half  quarters. 

In  this  description  we  have  supposed  the  rule  to  he 
folded ;  let  it  now  he  opened,  and  slide  out  the  slider, 
you  will  find  the  hack  part  of  it  divided  like  the  edge 
of  the  rule,  so  that  altogether  will  measure  1  yard  or 
3  feet  in  length. 

Some  rules  have  other  scales  and  tahles  upon  them ; 
as  a  table  of  hoard  measure,  one  of  timber  measure; 
a  line  for  shewing  what  length  for  any  breadth  will 
make  a  foot  square  ;  also  a  line  shewing  what  length 
for  any  thickness  will  make  a  solid  foot ;  the.  former 
Jine  serves  to  complete  the  table  of  hoard  measure,  and 
the  latter  the  table  of  timber  measure. 

The  thickness  of  a  rule  is  generally  about  a  quar- 
ter of  an  inch;  this  face  is  divided  into  inches,  and 
tenths,  and  numbered,  when  the  rule  is  opened,  from 
the  right-hand  towards  the  left,  10,  20,  30,  40,  &c. 
towards  100,  which  falls  upon  the  joint ;  the  other 
half  is  numbered  in  the  same  manner,  and  the  same 
way.  This  scale  serves  for  taking  dimensions,  in  feet, 
tenths,  and  hundredths  of  a  foot,  which  is  the  most 
commodious  way  of  taking  dimensions,  when  the  con- 
tents are  cast  up  decimally. 

The  use  of  the  Sliding  Rule. 

PROBLEM    I. 

To  multiply  Numbers  together,  as  12  and  16. 
Set  one   on  B.   to  the  multiplier  (12)  on  A;  then 
against  the  multiplicand  (16)  on  B,  stands  the  product 
(192)  on  A. 


Of  the  Carpenter's  Rule.  55 

II.  Find  the  product  of  35  and  19. 

Set  1  on  B,  to  the  multiplicand  (35)  on  A;  then  be- 
cause 19  on  B  runs  beyond  the  rule,  I  look  for  1.9  on 
B,  and  against  it  on  A,  I  find  66.5;  but  the  real  mul- 
tiplier was  divided  by  10,  therefore  the  product  68.5 
must  be  multiplied  by  10,  which  is  done  by  taking 
away  the  decimal  point,  so  the  product  is  665. 

PROBLEM    II. 

To  divide  one  Number  by  another,  as  360  by  12. 
Set  the  divisor  (12)  on  A,  to  1  on  B  ;  then  against 
the  dividend  (360)  on  A,  stands  the  quotient*  (30)  on  B. 

II.  Divide  7680  by  24. 
Set  the  divisor  (24)  on  A,  to  1  on  B;  then  because 
7680  is  not  contained  on  A,  I  look  for  768  on  A,  and 
against  it  I  find  32  on  B,  the  quotient;  but  because 
one-tenth  of  the  dividend  was  taken  to  make  it  fall 
within  the  compass  of  the  scale  A,  the  quotient  must 
be  multiplied  by  10,  which  gives  320. 

PROBLEM  III. 

To  Square  dny  Number,  as  25. 
Set  1  upon  C  to  10  upon  D — Then  observe,  that  if 
you  call  the  10  upon  1),  1,  the  1  on  C  will  be  1 ;  if 
you  call  the  10  on  D,  10,  then  the  1  on  C  will  be  100; 
if  you  call  the  10  on  D,  100,  then  the  1  on  C  will  be 
1000,  &c.  This  being  well  understood,  you  will  ob- 
serve that  against  every  number  on  D,  stands  its 
square  on  C. 

Thus  against  25-  stands     625 

against  30  stands     900 

against  35  stands  1225 

against  40  stands  1600 

Reckoning  the  10  on  D,  to  be  10. 


56  Of  the  Carpenter's  Rule. 

PROBLEM    IV. 

To  extract  the  Squart  Root  of  a  Number. 
Fix  (!)  •  slider  exactly  as  in  the  preceding  problem, 
and  estimate   ! Ik-  value   of  the  lines  D  and   C  in  the 
same  manner;  then  against  every  number  found  on  C 
stands  its  square  root  on  D. 

problem  v. 
To  find  a  mean  Proportional  between  two  given  Num- 
bers, as  9  and  25. 

Set  the  one  number  (9)  on  C,  to  the  same  (9)  on 
D;  then  against  25  on  C,  stands  15  on  D,  the  mean 
proportional  sought. 

For,  9  :  15    ::    15  :  25. 

2.  What  is  the  mean  proportional  between  29  and 
430  ? 

Set  29  on  C,  to  29  on  D  $  this  being  done,  you  will 
find  that  430  on  C  will  either  Fall  beyond  the  scaJe  D> 
or  it  will  not  be  contained  on  C.  Therefore  take  the 
100th  part  of  it,  and  look  for  4.3  on  C,  and  against  it 
on  1)  stands  11.2,  which  multiply  by  10,  and  112  is 
the  mean  proportional  required. 

PROBLEM   VI. 

To  find  a  fourth  Proportional  to  three  Numbers;  or, 
to  perform  the  Pule  of  Three. 

Suppose  it  were  required  to  find  a  fourth  propor- 
tional to  12,  28,  and  57. — Set  the  first  term  (12)  on  B 
to  the  second  (28)  on  A;  then  against  the  third  term 
(57)  on  B  stands  the  fourth  (133)  on  A. 

If  either  of  the  middle  numbers  fall  beyond  the  line, 
take  one-tenth  part  of  that  number,  and  increase  the 
answer  10  times. 

Note.  The  finding  a  third  proportional  between 
two  numbers  is  exactly  the  same,  for  in  such  cases 
the  second  number  is  repeated  to  form  the  third: 
Thus  the  third  proportion  to  12  and  28  is  65.33,  and 
is  thus  found  12  :  28  ::  28  :  G5.33.  viz.  set  12  on  B,  to 
29  on  A  5  then  against  28  on  B  stands  €5.3  on  A. 


Practical  Geometry*  Bt 


CHAPTER  XII. 

Practical  GeomeTrt. 

definitions. 

1.  GEOMETRY  is  a  science  which  teaches  and  cfe° 
monstrates  the  properties,  affections,  and  measures  of 
all  kinds  of  magnitude,  or  extension  5  as  solids,  sur- 
faces and  lines. 

2.  Geometry  is  divided  into  two  parts,  theoretical 
and  practical.  Theoretical  geometry  treats  of,  and  con- 
siders, the  various  properties  of  extension  abstract- 
edly ;  and  practical  geometry  applies  these  considera- 
tions to  the  various  purposes  of  life. 

3.  A  solid  is  a  figure,  or  body,  of  three 
dimensions,   viz.   length,    breadth    and 
thickness.     And  its  boundaries  are   su- 
perficies, or  surfaces.     Thus  A 
sents  a  solid. 

4.  A  superficies,  or  surface,  is  an  extension  of  two 
dimensions,  viz.  length  and  breadth,  with- 


out  thickness :    And   its  boundaries  are          B 
lines.     Thus  B  represents  a  surface.  

5.  A  line  is  length  without  breadth,  and  is  form- 
ed by   the   motion    of  a   point.     Thus ~- 

CD  represents  a  line.     Hence  the  ex-  C  D 

tremities  of  lines  are  points,  as  are  likewise  their  in- 
tersections. 

6.  Straight  lines  are  such  as  cannot  coincide  in  two 
points  without  coinciding  altogether,  and  a  straight 
line  is  the  shortest  distance  between  two  points. 

7.  A  point  is  position,  without  magnitude. 

8.  A  plain  rectilineal  angle  is  the 
inclination,  or  opening,  of  two  right 
lines  meeting  in  a  point,  as  E. 


58  Prt  nj. 

9.  One  angle  is  said  to  be  less  than  another,  when 
the  lines  which  form  it  are  nearer  to 
.  other.  Takfc  two  lines  AH  and 
BC  lunching  each  other  in  the  point 
B:  Conceive  these  two  lines  to  open, 
like  the  legs  of  a  pair  of  compasses, 
so  as  always  to  remain  fixed  to  each 
other  in  B. — While  the  extremity  A  moves  from  the 
extremity  C  the  greater  is  the  opening  or  angle  ABC; 
ami  on  the  contrary,  the  nearer  yon  bring  them  toge- 
ther, the  lt^s  the  opening  or  angle  will.be. 

to.  A  circle  is  a  plane  figure  eontain-        G>— -B 

fed  by  one  line  called  the  circumference,  J*         y. 

which  is  every  where    equally   distant  f      C       J 

from  a  point  within  it,  called  the  centre,  V  J 

as  C  :  And  an  arch  of  a  circle  is  any      ^»»  S 
part  of  its  circumference,  as  Gil. 

11.  The  magnitude  of  an  angle  does  not  consist  in 
the  length  of  the  lines  which  form  it,  but  in  their 
opening  or  inclination  to  each  other.  Thus  the  angle 
ABC  is  less  than  the  angle  DBE,  though  the  lines 
AB  and  CB,  which  form  the  former  angle,  are  longer 
than  the  lines  DB  and  BE,  which  form  the  latter. 

12.  When  an  angle  is  expressed  by 
three  let!  vBC,  the  middle 
letter  always  stands  at  the  angular 
point,  and  the  other  two  letters  at 
the  extremities  of  the  lines  which  B 
form  the  angle.  Thus  the  angle 
ABC  is  formed  by  the  lines  AB  and 
BC,  that  of  DBE  is  formed  by  the 
lines  DB  and  BE. 

13.  Every  angle  is  measured  by  an  arch  of  a  cir- 
cle, described  about  the  angular  point  as  a  centre ; 
thus  the  arch  1)E  is  the  measure  of  the  angle  DBE, 
and  the  arch  GH  is  the  measure  of  the  angle  ABC. 


Practical  Geometry.  5& 

14.  The  circumference  of  every  circle  is  supposed 
to  be  divided  into  360  equal  parts,  called  degrees  ; 
each  degree  into  60  equal  parts,  called  minutes;  and 
each  minute  into  60  equal  parts,  called  seconds. 

Angles  are  measured  by  the  number  of  degrees  cut 
off  from  the  circle,  by  the  lines  which  form  the  angles. 
Thus,  if  the  arch  GH  contain  20  degrees,  or  the 
eighteenth  part  of  the  circumference  of  the  circle, 
the  measure  of  the  angle  ABC  is  said  to  be  20  de- 
grees. 

15.  When  a  right  line  EC  stand- 
ing upon  a  right  line  AB,  makes 
the  adjacent  angles  ACE  and  BCE 
equal  to  each  other;  each  of  these 
angles  is  said  to  be  a  right  angle, 
and  the  line  EC  is  perpendicular  to 
AB.  The  measure  of  a  right  angle 
is  therefore  90  degrees,  or  the  quar- 
ter of  a  circle. 

16.  Jin  acute  angle  is  less  than  a  right  angle,  as 
DCB,  or  ECD. 

17.  Jin  obtuse  angle  is  greater  than  a  right  angle  as 
ACD. 

18.  A  plane  triangle  is  a  space  included  by  three 
straight  lines,  and  contains  three  angles. 

19.  A  right  angled  triangle  is  that  which 
lias  one  right  angle  in  it,  as  ABC.  The 
side  AC,  opposite  the  right  angle,  is  ealled 
the  hypothVnuse,  the  side  BC  is  called 
the  perpendicular,  and  the  line  AB,  on 
which  the  triangle  stands,  is  called  the 
base. 

20.  An  obtuse  angled  triangle  has  one 
obtuse  angle  in  it,  as  B. 


21.  An  acute  angled  triangle  has  all  its 
three  angles  acute,  as  C. 


M 


Practical  Geometry. 


is  lliat  which 
!ia^  three  equal  sides,  ami  three  equal  an- 
gles, as  I). 

An  isosceles  triangle  lias  two  equal 
sideband  the  third  side  either  greater  or 
less,  than  each  of  the  equal  sides,  as  E. 


24-.  A  scalene  triangle  has   all  its   three 
sides  uneqal,  as  F. 

2/5.  A  quadrilateral  figure  is  a  space  included  by 
four  straight  lines,  and  contains  four  angles. 

26,  A  parallelogram  is  a  plane  figure  hounded  by 
four  right  lines,  whereof  those  which  are  opposite  are 
parallel  one  to  the  other ;  that  is,  if  produced  ever  so 
far  would  never  meet. 


27.  A  square  is  an  equilateral  parallel- 
ogram, viz.  bavins  all  its  sides  equal,  and 
all  its  angles  riglft  angles,  as  G. 


28.  An  oblong  is  a  rectangled  parallel-  ^____ 
ogram,  whose  length  exceeds  its  breadth,  [       H 
asH.  ' ' 

29.  A  rhombus  is  a  parallelogram 
having  all  its  sides  equal,  but  its 
angles  are  not  right  angles,  as  I. 


30.  A  rhomboides  is  a  parallelogram 
having  its  opposite  sides  equal,  but  its       /"" 
length  exceeds  its  breadth,  and  its  an-    £•» 
gles  are  not  right  angles,  as  K. 


y 


31.  A  trapezium  is  a  plane  figure 
contained  under  four  right  lines,  no 
two  of  which  are  parallel  to  each  oth- 
er, as  L.  A  line  connecting  any  two 
opposite  angles  of  the  trapezium,  as 
AB.  is  called  a  diagonal. 


\ 


u 


Practical  Geometry, 


61 


32.  .#  trapezoid  is  a  plane  quadrilaler-  y        t 

al  figure,  having  two  of  its  opposite  sides      y^  %f    \ 
parallel,  and  the  remaining  two  not  par-   Z.  \ 

allel ;  as  M. 

83.  Multilateral  figures,  or  polygons,  have  more 
than  four  sides,  and  receive  particular  names,  accord- 
ing to  the  number  of  sides.  Thus,  a  Pentagon,  is  a 
Polygon  of  five  sides ;  a  H&cagon}  has  six  sides  ;  a 
Heptagon,  seven ;  an  Octagon,  eight ;  a  Nonagon, 
nine  5  a  Decagon,  ten ;  an  Undecagon,  eleven  ;  and  a 
Duodecagon  has  twelve  sides.  If  all  the  sides  and  an- 
gles are  equal,  they  are  called  regular  polygons  5  if 
unequal,  they  are  called  irregular  Polygons,  or  Fi- 
gures. 

34.  The  diameter  of  a  circle  is  a  right  line  drawn 
through  the  centre,  and  terminated  by  the  circumfer- 
ence both  ways  5  thus  AB  is  a  diameter  of  the  circle. 
The  diameter  divides  the  surface, 
and  circumference,  into  two  equal 
parts,  each  of  which  is  called  a 
semicircle.  If  a  line  CD  be  drawn 
from  the  centre  C  perpendicular  to 
AB,  to  cut  the  circumference  in  D, 
it  will  divide  the  semicircle  into  two 
equal  parts,  ACD  and  DCB,  each 
of  which  will  be  a  quadrant,  or  one-fourth  of  the  cir- 
cle. The  line  CD,  drawn  from  the  centre  to  the  cir- 
cumference, is  call  d  the  radius. 

35.  A  sector  of  the  circle  is  comprehended  under  two 
radii,  or  semidiameters,  which  are  supposed  not  to 
make  one  continued  line,  and  a  part  of  the  circum- 
ference. Hence  a  sector  may  be 
either  less  or  greater,  than  a  semi- 
circle 5  thus  ACB  is  a  sector  less 
than  a  semicircle,  and  the  re- 
maining part  of  the  circle  is  a  sec- 
tor greater  than  a  semicircle. 

36.  The   chord  of  an  arch  is  a 
right   line    less   than  the    diame- 

G 


&4  Practical  GLeometry. 

ter,  joining  flu-  extremities   of*  the  arch  ;  thus  DB  is 
the  chord  of  the  arch  DGE,  or  of  the  arch  EBAD. 

pneni  i*  any  part  of  a  circle  bounded  by  an 
arch  and  its  chord,  and  may  be  either  greater  or  less 
than  a  semicircle. 

.  Concentric  circles  are  those  which 
have  the  same  centre,  and  the  space 
included  between   their  circumferences 

lied  a  ring,  as  D. 


39.  If  two  pins  he  fixed  at  the  points  F,  /,  and  a 
thread  YPf  he  put  round  them  and  knotted  at  J*  : 
then  if  the  point    P   and   the  thread  he   moved  about 

the  fixed  centres  F,  /,   so   as  ^ ^ 

to    keep    the    thread    always 
stretched,  the  point  P  will  de- 
scribe   the     curve    PBDACP    A( 
called  an  ellipsis. 

40.  The  points  or  centres  F, 
jf.   are   called    the  foci,    and 

their  distance  from  C,  or  D?  is  equal  to  the  half  of  AB. 

4t.  The  line  AB,  drawn  through  the  foci  to  the 
curve,  is  called  the  transverse  axis,  or  diameter.  The 
point  O,  in  the  middle  of  the  axis  AB,  is  the  centre  ot 
the  ellipsis. 

The  line  CD,  drawn  through  the  centre  O,  per- 
pendicular to  the  transverse  diameter  AB,  is  called 
the  conjugate  axis,  or  diameter. 

43.  The  line  LR,  drawn  through  the  focus  F,  per- 
pendicular to  the  transverse  axis,  is  called  the  para- 
meter^ or  latus  rectum. 

ll.  A  line  drawn  from  any  point  of  the  curve,  per- 
pendicular to  the  transverse  axis,  is  called  an  ordinate 
to  the  transverse,  as  EG.  If  it  go  quite  through  the 
figure,  as  EH,  it  is  called  a  double  ordinate. 


Practical  Geometry* 


m 


45.  The  extremity  of  any  diameter  is  called  the 
vertex;  thus  the  vertices  of  the  transverse  diameter 
AB,  are  the  points  A  and  B. 

46.  That  part  of  the  diameter  between  the  vertex 
and  the  ordinate  is  called  an  abscissa;  thus  GB  and 
AG,  are  abscisses  to  the  ordinate  GE. 


47.  If  one  end  of  a  thread 
equal  in  length  to  CD,  be  fixed 
at  the  point  F,  and  the  other 
end  fixed  at  D,  the  end  of  the 
square  BCD  ;  and  the  side  CB 
of  the  square  be  moved  along 
the  right  line  AB,  so  as  always 
to  coincide  therewith,  the  string 
being  kept  stretched  and  close 
to  the  side  of  the  square  PD, 
the  point  P  will  describe  a  curve  HROLPG,  called  a 
parabola. 

4S.  The  fixed  point  F  rs  called  th£  focus. 

49.  The  right  line  AB  is  called  the  directrix. 

50.  The  line  ON  is  the  axis  of  the  parabola,  and 
O  is  the  vertex. 

51.  A  line  LR,  drawn  through  the  focus  F,  perpen- 
dicular to  the  axis,  is  called  the  parameter,  or  latus 
rectum. 

52.  A  right  line  IK,  drawn  from  the  curve  perpen- 
dicular to  the  axis,  or  parallel  to  the  directrix,  is  cal- 
led an  ordinate ;  if  it  go  quite  through  the  figure,  as 
IM  it  is  called  a  double  ordinate. 

53.  The  part  of  the  axis,  KO,  between  the  vertex, 
O,  and  ordinate,  IK,  is  called  the  abscissa. 


Note.  The  definitions  of  the  solids  are  given  in 
Part  II.  Chap.  II.  in  the  respective  sections  where 
each  solid  is  considered. 


64  Practical  Geometry. 


To  bisect,  or  divide  a  given  right  line  AB,  into  two 

From  the   points    A  and  B  with  any  ^.O* 

radius,   or    opening  of  the   compasses,       .    ?\ 
greater  than    half  AB,    describe    two 
arches  cutting  each  other  in  C  and  I)  ; 
draw  CD.  and  it   will   cut    AB    in  the 
point  E.  making  AE  equal  to  EB. 


7 


*-'£fr+ 


PROBLEM  II. 


At  a  given  distance  E,  to  draw  a 
right  line  CD  parallel  to  a  given  ",      ,       _         fl 
right  line  AB.  ■  '        ^       >. 

From  any   two  points  m  n  in  the         *»  /»"■"** 

line    AB    with  the  extent   of  E  in        R 
your  compasses  describe  two  arches  o 
s; — draw  CD  to  touch  these  arches,  without   cutting 
them,  and  it  will  be  parallel  as  required. 


PROBLEM    III. 

Through  a  given  point  o,  to  draw  a  right  line  CD,  par- 


allel to  a  given  straight  line  AB. 


Take  any  point  m  at  pleasure  in 
the  line  AB  5  with  m  as  a  centre 
and  distance  in  0,  describe  the  arch 
o  n  ;  with  n  as  a  centre  and  the 
same  distance  describe  the  arch  sm. 
Take  the  arch  on  in  your  compasses,  and  apply  it 
from  m  tost  through  o,s,  draw  CD,  and  it  will  be  par- 
allel as  required. 


££ 


Practical  Geometry, 


eg 


PROBLEM    IV. 


To  divide  a  given  right  line  AB  into  any  number  of 
equal  parts. 

From  the  ends  A  and  B  of  the  given  line,  draw  two 
lines  AP  and  BK  of  any 
length,  parallel  to  each 
other.  Then  set  any  num- 
ber of  equal  parts  from  A 
towards  P,  and  likewise  from 
B  towards  K  5  draw  lines  be- 
tween the  corresponding 
points  HM,  GL,  FT^  &e.  and  they  will  divide  AB  into 
the  equal  parts  AC,  CD,  DE  EB. 


PROBLEM    V. 


To  divide  a  given  rjght  line.  AB  into  two  such  parts, 
as  shall  be  to  each  other,  as  mr,  torn. 


From  A  draw  any  line  AC, 
equal  to  mn,  and  upon  it 
transfer  the  divisions  of  the 
line  mn,  viz.  mv  and  m. 
Join  BC,  and  parallel  to  it 
draw  DE;  then  will  AE  : 
EB  ::  mr:  rn. 


fly- 


PROBLEM    VI. 


To  find  a  third  Proportional  to  two  given  lines  AB,  AD. 


Place   the  two 


lines  B 


~D 


given 
and  AD  so  as  to  make  an  angle 
with  each  other  at  A;  in  AB 
the  greater,  cut  off'  apart  AC, 
equal  to  AD  the  less  given  line  ; 
join  BD,  and  draw  CE  parallel 
to  it,  then  will  AEbe  the  third  proportional  required, 
m.  AB  :  AD ::  AD  :  AE. 

G  2 


■ 


TROBLI.M    VII. 

To  find  a  fourth  Proportional  to  three 
AB,  AC,  AD. 


given  lines. 


Place  two  of  the  given  lines  AB  A., 
and  AC  so  as  to  make  an  angle  with  A- 
each  oilier  at  A,  and  join  BC.  On  ^" 
AB,  set  oft"  the  distance  AD,  and 
draw  DE  parallel  to  BC ;  then  AE 
is  the  fourth  proportional  required, 
viz.  AB :  AD  ::  AC  :  AE. 


A*r*T%n 


PROBLEM    VIII. 


From  a  given  point  V  in  a  right  line  AB  to  erect  a  Per- 
pendicular. 

1.  When  the  point  is  in,  or  near,  the  middle  of  the  line. 


On  opposite  sides  of  the  point  P 
take  two  equal  distances  P?n,  Pn 
from  the  points  m  and  nas  centres, 
with  any  opening  of  the  compass- 
es greater  than  Pw,  describe  two 
arches  cutting  each  other  in  r; 
through  r,  draw  CP,  and  it  will  be 
the  perpendicular  required. 


~?k 


2.  When  the  point  P  is  at  tlie  end  of  the  line. 


With  the  centre  P  and  any  ra- 
dius describe  the  arch  m  n  o  ;  set 
off  Pm  from  m  to  w,  and  from  n 
with  the  same  radius  describe  an 
archr:  through  m  and  n  draw 
the  line  m  n  r  to  cut  the  arch  in 
r;  then  through  r  and  P  draw 
CP  and  it  will  be  the  perpendicu- 
lar required. 


f  A 


Practical  Geometry. 

or  thus: 
Set  one  foot  of  the  compasses 
in  P  and  extend  the  other  to  a- 
ny  point  n,  out  of  the  Hue  AB  ; 
from  n    as    a   centre  and    dis- 
tance nP,   describe  a  circle  cut- 
ting AB   in  m  5  through    m  and  \ 
n9   draw  m  n  o  to  cut  the   circle     ^:i   \^ 
in  o,  then  through   o  draw  CP,  »v* 
which  will  be  perpendicular   to 
AB. 


PROBLEM    IX. 

From  a  given  point  C  to  let  fall  a  Perpendicular  up- 
on a  given  line  AB. 

1.  When  the  point  is  nearly  opposite  the  middle   of 

the  line. 

From  the    centre  C  describe  an  {J 

arch  to   cut  AB  in  m  and  n  ;  with 
the  centres  m  and  n,  and  the  same  ra- 
dius, describe  arches  intersecting  in    £\fq,      p    n/-& 
o  ;  through  C  and  o  draw  CP,   the  ***— 

perpendicular  required. 

2.  When  the  point  is  nearly  opposite  the  end  of  the  line, 

Frodpe  given  point  C  draw  any 
line  Cm  *  meeting  AB  in  the  point 
m;  besectraC  in  n,  and  with  n  as 
a  centre  and  radius  n  m,  or  wC  de-  » 
scribe  an  arch  cutting  AB  in  o, 
draw  Co,  and  it  will  be  the  perpendicular  required. 


Oh  Practical  Geometry. 


PROitLKM    x. 

In  any  Triangle  ABG  to  draw  a 
Perpendicular  from  any  tingle  h> 
t<»  it*  opposite  side. 

Bisect  either  of  the   sides  con- 
taining the  ankle  from  which  the 

perpendicular  is  to  be  drawn,  as 
BC  in  m  ;  with  the  radius  m  IJ  and 
centre  m  describe  an  areh  cutting 
AB  (produced  if  necessary)  in  I), 
draw  CI),  and  it  will  be  the  per- 
pendicular required. 

PROBLEM    XI. 

Upon  a  given  right  Une  AB  to  describe  an  Equilatera 
Triangle. 

With  B  as  a  centre  and  radius  equal 
to  AB  describe  an  arch  5  with  A  as  a 
centre  and  distance,  AB,  cross  it  in  C  ; 
draw  AC  and  BC,  then  will  ABC  be 
the  equilateral  triangle  required. 


PROBLEM    XII. 

To  make  a  Triangle  with  three  given  lines  AB,  A 
and  BC,  of  which  any  two  taken  together  are  gret 
than  the  third. 


With  A  as  a  centre  and  radius 
AC,  describe  an  arch,  with  the 
centre  B  and  distance  BC  cross 
it  in  C,  draw  AC,  and  BC,  then 
ABC  is  the  triangle  required. 


A 


Practical  Geometry.  $9 

PROBLEM   XIII. 

Two  sides  AB  and  BC  of  a  right 
angled  Triangle  are  given,  to 
find  the  Hypotlienuse. 

From  the  point  B  in  AB  draw 
BC    perpendicular   and    equal   to 
BC  :  join  AC,  and  it  will  be  the      A~"              -B 
hypotlienuse  required.  „ n 

PROBLEM    XIV. 

The  Hypotlienuse  AC,  and  one  side  AB,  of  a  right  an- 
gled Triangle  are  given,  to  find  the  other  side. 

Bisect  AC  in  m,  with  m  as  a 
centre  and  distance  m  A  describe 
an  arch,  with  A  as  a  centre  and 
distance  AB  cross  it  in  B:  join 
BC  :  then  ABC  is  a  right  angled 
triangle,  and  BC  the  required  side.  "7 


PROBLEM  xv. 

To  find  a  mean  Proportional  between  two  given  lines 
AB  and  BC. 
Join  AB  and  BC  in  one  straight    A  «* 


une  viz.  make  AC  rqual  to  the  sum    _    ~" 

yf  them,  and  bisect  it  in  the  point  o.   "         !£— ^.n 
With  the  centre  o  and  radius  o  A        /^y^k 
descrij^J^  semicircle ;   at  B   erect      Ax^^     j  >* 
the  pe^(Micular  BD,  and  it  will  be     Js        *?     k   d 
the  mean  proportional  required,  viz. 

AB  :  BD  ::  BD  :  BC. 
Note.  If  AD  and  DC  be  joined,  AD  will  he  a  mean 
proportional  between  AB  and  AC,  also  CD  will  be  a 
mean  proportional  between  BC  and  AC,  viz. 

AB  :  AD  ::  AD  :  AC 
and    BC  :  CD  :;  CD  :  AC. 


Practical  Geometry. 


PROBLEM    XVI. 

To  bisect,  or  divide  t  into  two  equal  parjs. 

From  the  centre C  With  any  radius 
describe  an  arch  win  from  m  and  u 
as  centres  with  the  game  radius  de 
scribe    two    arches    crossing 
other  in  o,  draw  Co,  and  it  will 
vide   the  given  angle  ACH  into  the  /w 
Ivvo  equal  angles  ACo  and  BCo. 


problem  xvli. 

At  a  given  point  A,  in  a  given  line  AB  to  make  an 
Jingle  equal  to  a  given  Angle  C. 

With  the  centre  C  and  any  radius 
describe  an  arch  m  n;  with  the  centre  A 
A  and  the  same  radius  describe  the 
arch  os:  Take  the  distance  mn  in 
your  compasses,  ana  apply  it  from  s  to 
o ;  then  a  line  drawn  from  A  through 
o  will  make  the  angle  A  equal  to  the 
angle  C. 


PROBLEM    XVIII. 

To  make  an  Angle  of  any  proposed  number  of  degrees 
upon  a  given  right  line,  by  the  Scale  of  Chords, 

Upon  the  line  AB  to  make  an 
angle  of  30  deg.  (Fig.  1.)  Take 
the  extent  of  (50  degrees  from 
the  line  of  chords  with  which 
and  the  centre  A  describe  the 
arch  am.  Take  30  degrees  from 
the  same  scale  of  chords,  and 
set  them  off  from  o  to  n  ;  through 
n  draw  An,  then  7?AB  is  the  an 
gle  required. 


Practical  Geometry.  iTi 

To  make  an  angle  of  150  degrees,  produce  the  line 
BA  to  m,  with  the  centre  A  and  the  chord, of  Go  de- 
grees ;  describe  a  semicircle  ;  take  the  given  obtuse 
angle  from  180  degrees,  and  set  off  the  remainder,  viz. 
30  degrees  from  m  to  n ;  through  n  draw  An,  then  n 
AB  is  the  angle  required. 

PROBLEM    XIX. 

*Sn  Angle  being  given  to  find  how  many  degrees  it  con- 
tains, by  a  Scale  of  Chords. 

With  the  chord  of  60  degrees  in  your  compasses  and 
centre  A  (Fig.  1.)  describe  the  arch  on,  cutting  the 
two  lines  which  contain  the  angle  in  o  and  n;  take 
the  distance  on  in  your  compasses,  and  setting  one 
foot  at  the  beginning  of  the  chords  on  your  scale,  ob- 
serve how  many  degrees  the  other  foot  reaches  to,  and 
that  will  be  the  number  of  degrees  contained  in  the 
arch  o  n,  or  angle  ftAB, 

If  the  extent  o  n  reach  beyond  the  scale,  which  will 
always  be  the  case  when  the  angle  is  obtuse,  produce 
the  line  BA  from  A  towards  m,  "and  measure  the  arch 
m  n  in  the  same  manner,  the  degrees  it  contains,  de- 
ducted from  ISO  degrees,  will  give  the  measure  of  the 
angle  nAB. 

PROBLEM    XX. 

Upon  a  given  right  line  AB,  to  describe  a  Square. 

With  A  as  a  centre  and  distance  &  £ 

AB  describe  the  arch  EB,  with  B 
as  a  centre  and  the  same  extent 
describe  the  arch  AC  cutting  the 
former  in  o  ;  make  o  E  equal  to 
B  o,  and  draw  BE,  make  oC  and 
ol)  each  equal  to  AF,  or  Fo,  and  A      "  H 

join  the  points  AD,  DC  and  CB,  then  will  ABCD  be 
the  square  required. 


Practical  Geometry. 


or  thus: 


Draw  BC  perpendicular,  and 
equal  to  AB  :  with  the  extent  AB 
and  one  foot  in  A  describe  an  arch; 
with  the  same  extent  and  one  foot 
in  C  cross  it  in  ]>.  join  AD  and 
DC,  then  ABCD  is  the  square  re- 
quired. 


PROBLEM  XXI. 


To  make  an  Oblong,  or  Rectangled  Parallelogram,  of  a 
given  length  BA,  and  breadth  BC. 


Place  BC  perpendicular  to  AB  ; 
with  the  centre  A  and  distance  BC 
describe  an  arch,  with  the  centre  C 
and  distance  AB  cross  it  in  D;  join 
AD  and  DC,  then  ABCD  is  the  ob- 
long required. 


it 


PROBLEM    XXII. 


Ujmn  a  given  right  line  AB  to  des- 
cribe a  Rhombus,  having  an  angle 
to  the  given  angle  A. 

Upon  AB  make  the  angle  DAB 
equal  to  the  given  angle  A;  also 
make  AD  equal  to  AB.  Then 
with  D  and  B  as  centres  and  radius 
AB,  describe  arches  crossing  each 
other  in  C  :  join  DC  and  BC,  then 
ABCD  is  the  rhombus  required. 


Practical  Geometry. 


£3 


PROBLEM  XXIII. 

To  find -the  centre  of  a  given  Circle 

Take  any  three  points.  A,  C,  B, 
in  the  circumference  of  the  circle, 
and  join  AC  and  BC  Bisect  AC 
and  BC  with  the  lines  no,  and  mo, 
meeting  each  other  in  the  point  o. 
Then  o  is  the  centre  required. 

Note.  By  this  problem  it  will\e  easy  to  describe 
a  circle  through  any  three  given  points  that  are  not  in 
the  same  straight  line  :  Or,  to  describe  a  circle  about. 
a  given  triangle.  x\lso,  if  any  areh  of  a  circle  be 
given,  the  whole  circle  may  be  readily  described. 


PROBLEM   XXIV. 

To  draw  a  right  line  equal   to  any  given  arch   of  a 
Circle,  A  B. 

Divide  the   chord  AB    into   four 
equal    parts ;  set  one   part    AC  on 
the  arch  from  B  to  D:  draw   CD, 
and  it  will  be  nearly  equal  to  half  the   length 
arch  ADB. 

OR   THUS : 

Through  the  point  A,  and  o 
the  centre  of  the  circle  draw  A 
m  ;  divide  o  n  into  four  equal 
parts,  and  set  off  three  of  them 
from  m  to  n.  Draw  AC  perpen- 
dicular to  A  m,  then  through  m 
and  B  draw  m  C ;  then  will  ilC 
be  equal  to  the  length  of  the 
arch  AB  very  nearly. 


H 


Practical  Geometry. 


PROBLEM    XXV. 

To  make  a  square  equal  in  Jlrea  to  a  given  circU, 


Divide  the  diameter  AB  into 
fourteen  equal  parts,  and  make. 
AX!  equal  to  eleven  of  these 
parts;  erect  the  perpendicular 
GC,  and  join  AC  ;  then  the 
square  AEDC,  formed  upon 
AC,  is  equal  to  the  whole  cir- 
cle whose  diameter  is  AB,  ex- 
ceedingly near  the  truth. 


PROBLEM    XXV] 


In  a  given  circle  to  describe  a  Sana  re. 


Draw  any  two  diameters  AB  and 
CD  perpendicular  to  each  other, 
then  connect  their  extremities,  and 
that  will  give  the  inscribed  square  A 
DBC. 


Note.  If  a  side  of  the  square,  as  DB,  be  bisected 
in  m,  and  aline  om  n  be  drawn  from  the  centre  of  the 
circle  to  cut  the  circumference:  then  if  the  line  Bn  be 
drawn,  it  will  be  the  side  of  an  octagon  inscribed  in 
the  circle. 

PROBLEM    XXVII. 

To  make  a  regular  Folygon  on  a  given  line  AB. 

Divide  360  degrees  by  the  num-  p 

her  of  sides  contained  in  your  poly- 
gon, subtract  the  quotient  from  ^ 
ISO  degrees,  and  the  remainder 
will  be  the  number  of  degrees  in 
each  angle  of  the  polygon.  From 
each  end  of    AB  draw  lines  AG, 


BO.    makini 


angles 


with     the 


gitenline  equal  to  half  the  angle 

o£  the  polygon:  then  with  the   centre  0  and   radius 


Practical  Geometry, 


iu 


©A  describe  a  circle  to  the  circumference  of  which  ap- 
ply continually  the  given  side  AB. 

or  thus: 
Take  the  given  line  AB  from  any  scale  of  equal 
parts  :  multiply  the  side  of  your  polygon  by  the 
number  in  the  third  column  of  the  following  table,  an- 
swering to  the  given  number  of  sides,  and  the  product 
will  give  you  the  length  of  AO  or  OB,  with  which  pro- 
ceed as  above. 


M.  of 

JVame  of  the 

Had.  of  the  circum- 

An. OAI3, 

or 

Sides. 

Polygon. 

scribing-  c- 

OBA. 

3 

Trigon, 

.5773503 

30 

4 

Tetragon, 

.7071068 

45 

5 

Pentagon, 

.8506508 

54 

b 

6 

Hexagon, 

1,  Side=Radius. 

60 

| 

7 

Heptagon, 

1.1523825 

64? 

8 

Octagon, 

1.3065630 

67]; 

s 

9 

Nocagon, 

1.4619022 

70 

10 

Decagon, 

1.6186340 

72  i 

11 

Undecagon, 

1.7747329 

73t\ 

12 

Duodecagon, 

1.9318516 

75 

PROBLEM   XXVIII* 

In  a  given  circle  to  inscribe  any  regular  Polygon  $  or  to 
divide  the  circumference  of  a  given  circle  into  any 
number  of  equal  parts. 

Divide  the  diameter  AB 
into  as  many  equal  parts  as 
the  figure  has .  sides  ;  from 
the  centre  o  draw  the  per- 
pendicular om  divide  the  ra- 
dius on  into  four  equal  parts 
and  set  off  three  of  these 
parts  from  n  to  m,  from  m, 
through  the  second  division 
s9  of  the  diameter  AB  draw 
mC  :  join  AC,  and  it  will  be 
the  side  of  the  polygon  re- 
quired. 


Practical  Geometry. 


PROBLEM    XXIX. 

the  two  diameters  of  an  Oval,  a  figure  re~ 
semblins  the  conic  section  called  an  Ellipsis,  to 
describe  it. 

Bisect  the  longer  diameter 
AB  in  the  point  C  by  the  line 
<wt  w,  bisect  the  shorter  diame- 
ter HI  in  K,setoffHKor  IK 
from  C  towards  m  and  n : 
then  from  A  set  off  \o  equal 
to  lit,  divide  the  remaining 
part  oB  into  three  equal  parts, 
and  set  off  two  of  them  from 
o  to  85  with  S  as  a  centre,  and 

radius  SS,  cross  the  line  m  n,  in  m  and  «,  through  S 
draw  m  SE,  m  SD,  #SG,  and  nSF,  of  an  indefinite 
length. — With  the  radius  AS  and  centre  S  describe 
the  arches  EAG  and  FBD  :  also  with  n  as  a  centre 
and  radius  nO  describe  the  arch  GF,  passing  through 
the  extremity  of  the  shorter  diameter,  and  meeting 
BF  in  the  point  F;  in  a  similar  manner,  with  m  as 
a  centre,  and  radius  711E  onnD,  describe  the  arch 
ED*,  which  will  complete  the  oval  required. 

Note.     There  are  various  methods  of  constructing 
lipsis,  and  oval  ;  the  former  may  be  accurately 
constructed  by  the  39th  and  40th  definition,  and  the 
latter  by  the  above  method.     It  may   be  proper  here 
to  inform  the  student,  unacquainted    with   this    sub- 
ject, that  it  is  impossible  to  describe  an  ellipsis  tru- 
ly with  a  pair   of  compasses;  for  an  ellipsis  has  no 
of  the   curve  of  a  circle  in   its  composition,  an 
tical  curve,  being  described  tm   two    points   (see 
:0.)  is  continually  varying. 


Practical  Geometry, 


97 


PROBLEM    XXX. 


Given   the  Jlb^cissa  vm,  and  ordinate  Sm,  of  a  para- 


bola* to  construct  it, 


Bisect  the  given  ordinate  Sm 
in  B  5  join  By  and  draw  BD 
perpendicular  to  it,  meeting  the 
abscissa  produced  in  D. 

Make  ov  and  vC9  each  equal 
to  Dm,  then  will  o  be  the  focus 
of  the  parabola. 

Take  any  number  of  points 
m.m.&c.  in  the  abscissa,  through 
which  draw  the  double  ordi- 
nates  S/uS,  &c.  of  an  indefinite 

length.  With  the  radii  oC,  mC,  &c.  and  centre  o,  de- 
scribe arches  cutting  the  corresponding  ordinates  in 
(he  points  S,  S,  &e.  and  the  curve  SvS,  drawn  through 
all  the  points  will  be  the  parabola  required. 


COMPLETE  MEASUJ1I/K 


PART  II. 


CHAPTER  I. 

•MENsuajfioN  of  Superficies. 

TtlE  area,  or  superficies,  of  any  plane  figure,  is  es- 
timated by  the  number  of  squares  contained  in  its  sur- 
face ;  the  side  of  those  squares  being  either  an  inch., 
a  foot,  a  yard,  a  link,  a  chain,  &c.  And  hence  the 
area  is  said  to  be  so  many  square  inches,  square  feet* 
■square  yards,  square  links  or  square  chains,  &c.  Our 
common  measures  of  length  are  given  in  the  first  table 
below,  and  the  second  table  of  square  measure  is  taken 
from  it,  by  squaring  the  several  numbers. 


Lineal  Measur 


12     Inches 

a     Feet 

6     Feet 
163-  Feet    > 

6\  Yards  5 
40  Poles 

8  Furlongs- 


Y 


Foot 
Yard 
Fathom 
Pole  or 

Rod 
Furlong 
Mile    ^ 


II. 

Square  Measure. 


144 


Inches 
9    Feet 
36    Feet 
27  2\  Feet 

30-1-  Yards 
1600    Poles 
64  Furlongs 


1  Foot 
1  Yard 
1  Fathom 
1  Pole  or 

Rod 
1  Furlong 
1  Mile 


•Land  is  measured  by  a  chain,  called  Gunter's  chain, 
«f  4  poles,  or  22  yards  in  length,  and  consists. of  100 
equal  links,  each  link  being  T%%  of  a  yard  in  length, 
<#r  7.92  inches.    Ten  square  drains,  or  ten  chains  in 


Mensuration  of  Superficies. 


?9 


length  and  one  in  breadth,  make  an  acre;  or  4840 
square  yards,  160  square  poles,  or  100,000  square 
links,  each  being  the  same  in  quantity.  Forty  perches, 
or  square  poles,  make  a  rood,  and  4  roods  make  an 
acre. 

The  length  of  lines  measured  with  a  chain,  are  ge- 
nerally set  down  in  links,  as  whole  numbers:  every 
chain  being  100  links  in  length.  Therefore  after  the 
dimensions  are  squared,  or  the  superficies  is  found,  it 
will  be  in  square  links ;  when  this  is  the  ease,  it  will 
be  necessary  to  cut  off  five  of  the  figures  on  the  right 
hand  for  decimals,  and  the  rest  will  be  acres.  These 
decimals  must  be  then  multiplied  by  4  for  roods,  and 
(lie  decimals  of  these  again,  after  five  figures  are  cut. 
•iff,  by  40  for  perches. 


§  I.  To  find  the  JIjrea  of  a  Square. 

Example  1.  Let  ABCD  be  a  square  given,  each  sida 
feeing  14.  Required  the  area,  or  superficial  content, 


Is£ 


14X14=196,  the  area  of  the  square  ABCD, 


Sfi  Mensuration  of  Superficies, 


By  Scale  and  Conqmsses. 

Extend  the  compasses  from  1,  in  the  line  of  num- 
bers, to  14;  the  same  extent  will  reach  from  the  same 
point,  turned  forward,  to  106.  Or,  extend  from  10  to 
14,  that  extent  will  reach  to  19.6,  which  multiply 

by  io. 

Demonstration.  Let  each  side  of  the  given  square 
he  divided  into  14  equal  parts,  and  lines  drawn  from 
one  another  crossing  each  other  within  the  square  ;  so 
shall  the  whole  great  square  lie  divided  into  196  little 
squares,  as  you  may  see  in  the  figure,  equal  to  the 
number  of  square  feet,  yards,  poles,  or  other  measure, 
by  which  the  side  was  measured. 


2.  What  is  the  area  of  a  square  whose  side  is  35.25 
chains  ?  Ans.  124  acres  1  rood  1  perch. 

3.  Required  the  area  of  a  square  whose  side  is   5 
feet  9  inches.  Jliis.  33  feet  0  inches  9  parts. 

4.  What  is  the  area  of  a  square  whose  side  is  3723 
links  ?  Jlns.  138  acres  3  roods  1  perch. 


§  II.  Examples  of  an  Obloxg,  or  rectangled  PARAL- 
LELOGRAM. 

To  find  the  area  of  a  parallelogram;  whether  it  he 
a  square,  a  rectangle,  a  rhombus,  or  a  rhomboid. 


Multiply  the  length  by  the  breadth,  or  perpendicu- 
lar height,  and  the  product  will  be  the  area. 


Mensuration  of  Superficies* 


Si 


0 

l£ 

B 

j 

1 

1 

~T~ 

T 

9 

- 

] 

\ 

1 

1> 

.    J 

1 

c 

Example  1.  Let  ABCD  be  an  oblong,  the  length  of 
it  18  feet,  and  the  breadth  9  feet;  these  multiply  to- 
gether, the  product  is  162,  the  superficial  content. 

By  Scale  and  Compasses. 

Extend  the  compasses  in  the  line  of  numbers  from 
18  to  162,  the  square  feet.  Or,  call  the  middle  unit  on 
the  scale  10,  and  extend  from  it  to  9,  that  extent  will 
reach  from  18  to  16.2,  which  multiply  by  10. 

Demonstration.  If  the  sides  AB  and  CD  he  eaeU 
divided  into  18  equal  parts,  representing  18  feet;  and 
the  lines  AD  and  BC  each  divided  into  9  equal  parts, 
and  lines  drawn  from  point  to  point,  crossing  each 
other  within  the  figure ;  those  lines  will  make  thereby 
so  many  little  squares  as  there  are  square  feet,  viz.  ±Q2. 

2.  What  is  the  superficial  content  of  a  rectangular 
hoard,  whose  length  is  14  feet  6  inches,  and  breadth 
4  feet  9  inches. 

Jns.  68  feet  10  inches  6  parts. 

3.  Required  the  area  of  a  rectangular  piece  of 
ground,  whose  length  is  1375  links,  and  breadth  950. 

Jfris,  13  acres  0  roods  10  perches. 

4.  What  is  the  superficial  eontent  of  an  oblong  which 
is  2  feet  10  inches  6  parts  long,  and  9  inches  broad  ? 

tins.  2  feet  1  inch  10  parts  6  second's. 


Mensuration  of  Superfic. 
§  HI.  Examples  of  a  RhombI's. 


Example  i.  Let  ABCD  be  a  rhombus  given,  whose 
sides  are  each  13.5  feet,  and  the  perpendicular  BA  is 
13.42  5  these  multiplied  together,  the  product  is 
208.010 ;  which  is  the  superficial  content  of  the  rhom- 
bus, that  is,  208  feet  and  one  hundredth  part  of  a  foot. 

By  Scale  and  Compasses. 

Extend  the  compasses  from  1  to  13.42,  that  extent 
will  reach  from  15.3  the  same  way  to  20Sfeet,  the  con- 
tent. Or,  call  the  middle  1  upon  the  scale  10  and  ex- 
tend from  it  to  13.42,  that  extent  will  reach  from  15.5 
to  20.8,  which  multiply  by  10. 

Demonstration.  Let  CD  be  extended  out  to  F, 
making  DF  equal  to  CE,  and  draw  the  line  BF;  so 
shall  the  triangle  DBF  be  equal  to  the  triangle  ACE: 
For  DF  and  GE  are  equal,  and  BF  is  equal  to  AE, 
because  AB  and  CF  are  parallel.  Therefore  the  par- 
allelogram ABEF  is  equal  to  the  rhombus  ABCD. 

2.  "VVhat  is  the  area  of  a  rhombus,  whose  length  is 
6.2  chains  and  perpendicular  height  5.45  chains? 

Jlns.  3  acres  1  rood  20.64  perches. 
The  length  of  a  rhombus  is  12  feet  6  inches,  and 
perpendicular  height  9  feet  7  inches,  what  is  the  area? 
Jlns.  119  feet  9  inehes  6  parts. 

4.  The  length  of  a  rhombus  is  725  links,  and  per- 
pendicular height  635  links,  what  is  the  area  ? 

Jlns.  4  acres  2  roods  16.6  perches. 


Mensuration  of  Superficies. 
§  IV.  Examfles  of  a  Rhomboides. 


irv- 


Ex ample.  1.  Let  ABCD  be  a  rhomboides  given*, 
whose  longest  side  AB  or  CD,  is  19.5  feet,  and  the 
perpendicular  AE  is  10.2;  these  multiplied  together, 
the  product  is  198.9,  that  is,  198  superficial  feet  and 
9  tenth  parts,  the  content. 

Demonstration.  If  DC  be  extended  to  F,  making 
CF  equal  to  DE,  and  a  line  drawn  from  B  to  F ;  so 
will  the  triangle  CBF  be  equal  to  the  triangle  ADE, 
and  the  parallelogram  AEFB  be  equal  to  the  rhom- 
hoides  ABCD ;  which  was  to  be  proved. 

2.  A  piece  of  ground,  in  th^ibrm  of  a  rhomboides, 
measures  4784-  links  in  length,^and  its  perpendicular 
breadth  is  1908  links,  what  is  its  area  ? 

•Sns.  91  acres  1  rood  4.5952  perches. 

3.  Required  the  area  of  a  rhomboides,  whose  length 
is  12  feet  6  inches,  and  perpendicular  breadth  5  feet 
0  inches.  Jlns.  68  feet  9  inches. 

4.  Plow  many  square  yards  of  painting  are  in  a 
rhomboides,  whose  length  is  37  feet  and  breadth  5  feet 
3  inches?  Jlns.  21  f2  yards. 


84 


Mensuration  of  Superficies. 


§  V.  To  find  the  JIrea  of  a  Triangle  when  the 
base  and  perpendicular  are  given. 


RULE. 


Multiply  the  base  by  the  perpendicular,  and  half 
he  product  will  be  the  area. 


EXAMPLES. 

a 
0 


1.  Let  ABC  he 
right  angled  triangle, 
whose  base  is  11.1 
and  the  perpend ieular 
12  feet.  Multiply  1  1.1 
by  6,  half  the  perpendi- 
cular, and  the  product 
is  84.6  feet,  (he  area. 
Or  multiply  14.1  by  12, 
the  product  is  169.2; 
the  half  of  which  is 
S4.6,  the  same  as  before. 


By  Scale  and  Compasses. 
Extend  the  compasses  from  1  to  14.1;  that  extent 
will  reach  the  same  way  from  6  to  84.6  feet,  tfee  con- 
tent. Or,  call  the  middle  1  on  the  scale  10,  extend 
from  it  to  1 4.1,  that  extent  will  reach  from  6  to  8.46, 
which  multiply  by  10. 

-$. S * 


Note.  The  per- 
pendicular in  the 
figure  ought  to  be 

7.8. 


2.  Let  ABC  (Fig.  2.)  be  an  oblique-angled  trian- 
gle given,  whose  base  is  15.4,  and  the  perpendi- 
cular 7.8  ;  if  15.4  be  multiplied  by  3.9  (half  the 
perpendicular,)   the  product  will  he  60.06  for   the 


Mensuration  of  Superficies,  8o  . 

area,  or  superficial  content :  Or,  if  the  perpendicular 
7.8  be  multiplied  into  half  the  base  7.7,  the  product 
will  be  90.06  as  before:  Or,  if  15.4,  the -base,  be  mul- 
tiplied by  the  whole  perpendicular  7.8,  the  product 
will  be  120.12,  which  is  the  double  area;  the  half 
whereof  is  60.06  feet  as  before. 


By  Scale  and  Compasses. 

Extend  the  compasses  from  2  to  15.4,  that  extent 
will  reaeh  from  7.8  to  60  feet,  the  content.  Or,  extend 
from  20  to  15.4,  that  extent  will  reach  from  7.8  to  6, 
which  multiply  by  10. — Here  the  middle  1  on  the 
scale  is  considered  as  10. 

Demonstration.  If  AD  (Fig.  1.)  be  drawn  parallel 
to  BC,  and  DC  parallel  to  AB ;  the  triangle  ADC 
shall  be  equal  to  the  given  triangle  ABC.  Hence  the 
parallelogram  ABCD  is  double  of  the  given  triangle ; 
therefor-  half  the  area  of  the  parallelogram  is  the  area 
of  the  triangle.  In  Fig.  2.  the  parallelogram  ABEF 
is  also  double  of  the  triangle  ABC;  for  the  triangle 
ACF  is  equal  to  the  triangle  ACD,  and  the  triangle 
BCE  is  equal  to  the  triangle  BCD ;  therefore  the  area 
of  the  parallelogram  is  double  the  area  of  the  given 
triangle :  Which  was  to  be  proved. 

3.  The  base  of  a  triangle  is  28.2  yards,  and  the  per- 
pendicular height  18.4  yards,  what  is  the  area? 

Jlns.  259.44<  square  yards. 

4.  There  is  a  triangular  field  whose  base  measures 
1236  links,  and  perpendicular  731  links;  how  many 
acres  does  it  contain  ? 

Ans.  4  acres  2  roods  2.8128  perches. 

5.  What  is  the  area  of  a  triangle  whose  base  is  18 
feet  4  inches,  and  perpendicular  height  11  feet  10 
inches  ?  Jlns*  10S  feet  5  inches  8  parts. 

I 


&6-  Mi  a  8  u  rat  ion  of  Superfi  c  ie& , 


To  find  the  area  of  any  plane  Triangle  by  having  the 
three  sides  given,  without  the  help  of  a  Perpendi- 
cular. 

ruli:. 

Add  the  three  sides  together,  and  take  half  that 
sum;  from  which  stibstract  each  side  severally;  then 
multiply  the  half  sum  and  the  three  differences  con- 
tinually, and  out  of  the  last  product  extract  the  square 
root  5  which  will  be  the  area  of  the  triangle  sought. 

EXAMPLES. 

!•  Let  ABC  be  a  triangle,  whose  three  sides  are 
as  follows;  viz.  AB  43.3.  AC  20.5,  and  BC  31.2, 
the  area  is  required. 


("43.3  |    4.2") 

1  31.2      16.3  I  Di 

[20.5  J  27.0  J 


Sides  <  31.2    ir>.3  >.  Differences. 


Sum   95.0 
Half  47.5 


Mensuration  of  Superficies, 


*7 


Area  296.31 

47.5  The  half  sum. 
27  Difference* 

3325 
950 

1282.5  Product. 
16.3  Difference. 

3S4Jb 

76950* 
12825 

20904.75  Product. 
4.2  Difference. 

4180950 
S361900 

87799.950  Last  product 

the  square  root  of  which  is  296.31  (remainder  3339) 
the  area  required. 


2.  What  is  the  area  of  a  triangle  whose  sides  are 


50,  40  and  30  ? 


Jins.  600. 


'    3.  The  sides  of  a  triangular  garden  are  41,  29  and 
56  yards,  what  is  the  area  thereof? 

Jins,  574.34  yards. 

4.  How  many  acres  are  contained  in  a  triangle  whose 
three  sides  are  4900,  5025  and  2569  links. 

Jins.  61  acres  1  rood  39.68  perches. 

5.  A  field  of  a  triangular  form,  whose  sides  are  3S0, 
420  and  765  yards,  rents  for  55  shillings  per  acre  j 
what  is  the  annual  rent  ? 

Jins.  The  area  is  44699.0347  yards,  or  9,235337 
acres,  which  at  55  shillings  per  acre  amount*  to 
1.25  :  7  :  11^28,  the  annual  rent. 


$s 


iitioU  of  >*- 


Demonstration  of  the  r.ule.— -In  the  triangle 
BDC,  I  say,  if  from  the  half  sum  of  Ihc  sides,  you 
subtract  each  particular 
.side,  and  multiply  the 
half  sum  and  the  three 
differences  together  con- 
tinually, the  square  root 
of  the  product  shall  be 
the  area  or  the  triangle. 

First,  by  the  lines  Bl, 
CI,  and  l)i,  bisect  the, 
thro.  which  lines 

will  all  meet  in  the  point 
I:  by  which  lines  the 
given  triangle  is  divided 
into  three  new  triangles 
CBI,  DCI,  and  BDI;  the  perpendiculars  of  which 
new  triangles  are  the  lines  AI,  EI,  and  OT,  being  all 
equal  to  one  another,  because  the  point  1  is  the  centre 
of  the  inscribed  circle  (Eutlid,  Lib.  IV.  Vrob..  4.) : 
Wherefore  to  the  side  BC  join  CF  equal  to  DE,  or 
DO;  so  shall  BF  be  equal  to  half  the  sum  of  the 
sitles  :  vix.  =  4  BC  +  *  BD  4.  -V  CD. 

AndBA—BF— CD;  for  CA=CO  and  OD—CF ; 
therefore  CD—AF;  and  AC— BF—BD,  for  BE= 
BA,  and  ED—CF;  therefore  BD^BA+CF,  and 
CF=BF— BC. 

Make  FH  perpendicular  to  FB,  and  produce  BI  to 
meet  it  in  II.  Draw  C1I  and  HK.  perpendicular  to  CD. 
Because  the  angles  FCK-f-EHK  are  equal  to  two 
right  angles  (fbr  the  .angles  F  and  K  are  right  angles) 
equal  also  to  FCK+ACO  (by  Euclid,  I.  13.)  and  the 
angles  ACOxAIO  are  equal  to  two  right  angles; 
therefore  the  quadrangles  FCKH  and  AIOC  are  simi- 
lar; and  the  triangles  Oil  and  AIC  are  also  similar. 
And  the  triangles  BAI  and  BFH  are  likewise  similar. 

From  this  explanation,  I  say,  the  square  of  the  area 
of  the  given  triangle:  that  is,  BF2"xlA3~BF 
XBAxCAxCF.    In  words: 


Mensuration  of  Superficies.  89 

The  square  of  BF  the  half  sum  of  the  sides  mul- 
tiplied into  the  square  of  IA  (—1  E— I  O)  will  be 
equal  to  the  said  half  sum  multiplied  into  all  the  three 
differences. 

ForIA:  BA  ::  FH:BF;andIA:  CF  :  :  AC  : 
FH  5  because  the  triangles  are  similar.  By  Euclid ', 
Lib.  VI.  Prop.  4. 

Wherefore  multiplying  the  extremes  and  means 
In  both,  it  vvillbe  IJ^+BFxFH=BAxCAxCF-{- 
FH ;  but  FH  being  on  both  sides  of  the  equation,  it 
may  be  rejected  ;  and  then  multiply  each  part  by  BF, 
it  will  be  "BT^xiA^zrBFxBAxCAxCF.  Which 
was  to  be  demonstrated. 

Jf  any  two  sides  of  a  right  angled  Triangle   be  gi~ 
:en,  the  third  side  may  be  found  by  the  following 


Pi 


RULE. 


1.  To  the  square  of  the  base  add  the  square  of  the 
perpendicular,  the  square  root  of  the  sum  will  give 
the  hypothenuse,  or  longest  side  ;  or  from  the  square 
of  the  hypothenuse  take  the  square  of  either  side,  and 
the  square  root  of  the  remainder,  will  be  the  other  side. 

Example  1.  Given  the  base  AB  (see  Fig.  1.  §  V.) 
14.1,  the  perpendicular  BC  12,  what  is  the  length  of 
the  hypothenuse  AC  ? 

14.1  the  base  AB.  12  the  perpend.  BC* 

14.1  12 

ill  144  square  of  BC. 

564 
ill 


198.81  square  of  AB 
144.      square  of  BC. 

3*2.81  Sum  of  the  squares,  the  square  root  of  which 
extracted  gives  AC  18.51513. 


90  Mensuration  of  Superjl 

Gtiven  the  base    *'0,  and  the  hypothenuse  2j  ; 
to  find  the  perpendicular. 

From  the    square  of    25— 
Take  the  square   of   15z_;  - 

-And  the    square  root  of     400 
will  be  the  pei  pendieular—^O. 

i'iie   base  of  a  right  angled  triangle  is  24,  and 
the  perpendicular  18.  what  is  the  hypothenuse  ? 

Ms.  30. 

4.  The  wall  of  a  fort  standing  on  the  brink  of  a  ri- 
ver is  42.426  feet  high,  the  breadth  of  the  river  is  23 
yards  :  what  length  must  a  eord  be  to  reach  from  the 
top  of  the  fort  across  the  river?  Ms.  27  yards. 

5.  The  hypothenuse  of  a  right  angled  triangle  is  30, 
and  the  perpendicular  IS,  what  is  the  base  ? 

Ms.  21. 

6.  A  ladder,  50  feet  long,  will  reach  to  a  window 
.JO  feet  from  the  ground  on  one  side  of  a  street ;  and 
without  moving  the  foot  will  reach  a  window  40  feet 
high  on  the  other  side.  The  breadth  of  the  street  is  re- 
quired. Ms.  2Z\  yards. 

7.  A  line  of  380  feet  will  reach  from  the  top  of  a 
precipice,  which  stands  close  by  the  side  of  a  brook,  to 
the  opposite  bank,  and  the  precipice  is  128  feet  high, 
how  broad  is  the  brook  ?  Ans.  357.29  feet. 

8.  If  a  ladder,  50  feet  in  length,  exactly  reach  the  co- 
ping of  a  house,  when  the  foot  is  10  feet  from  the  up- 
right of  the  building ;  how  long  must  a  ladder  be  to 
reach  the  bottom  of  the  second  floor  window,  which  is 
17.9897  feet  from  the  coping,  the  foot  of  this  ladder 


Mensuration  of  Superficies. 


01 


and 


standing  6  feet  from  the  upright  of  the  building ; 
what  is  the  he£$ht  of  the  wall  of  the  house  ? 

JLns.  The  height  of  the  wall  is  48.9897  feet,  and 
the  length  of  a  ladder  to  reach  the  second  floor  win- 
dow must  be  31.5753  feet. 


§  VI.  To  find  the  Area  of  a  Trapeziu. 


rule. 


Add  the  two  perpendiculars  together,  and  take  half 
the  sum,  which  multiply  by  the  diagonal :  the  product 
is  the  area.  Or,  find  the  areas  of  the  two  triangles, 
ABC  and  ACD  (by  section  V.)  and  add  them  toge- 
ther, the  sum  shall  be  the  area  of  the  trapezium. 


Mensuration  of  Superjl 

EXAMPLES. 

i.  Let  ABCD  be  a  trapezium  given,  the.  diagonal 
of  which  is  80.0,  and  the  perpendicular  13F  30.6j 
and  ilie  perpendicular  DBJ,  34.0  :  these  two  added 
together,  the  sum  is  54.6,  the  half  of  which  i- 
and  this  being  multiplied  by  the  diagonal,  80.3,  the 
product  is  2197.63,  which  is  the  area  of  the  trape- 
zium. 

By  Scale  and  Compasses. 

Extend  the  compasses  from  2  to  51.6 ;  that  extent 
will  reach  from  80.5  to  2197.65,  the  area.  Or  call  the 
unit  at  the  beginning  of  the  scale  100,  and  extend 
from  200  to  316,  that  extent  will  reach  from  S05  to 
2197.65. 

Demonstration.  This  figure  ABCD  is  composed 
of  two  triangles,  the  triangle  ABC  is  half  the  paral- 
lelogram AGHC:  also  the  triangle  ACD  is  equal  to 
half  the  parallelogram  ACIK,  as  was  proved,  sect.  V. 
Wherefore  the  trapezium  ABCD  is  equal  to  half  the 
parallelogram  GH1K.  To  find  the  area,  HI=BF-f- 
DE;  therefore! HI x AC  (=KI=GII)  area  of  the 
trapezium,  which  was  to  be  proved. 

2.  There  is  afield  in  the  form  of  a  trapezium,  whose 
diagonal  is  1660  links,  the  perpendiculars  702  and  713 
links, what  is  the  area? 

Ms.  11  acres  2  roods  37.792  perches. 

3.  What  is  the  area  of  a  trapezium  whose  diago- 
nal is  34  feet  9  inches,  and  the  two  perpendiculars  19 
feet  9  inches  and  8  feet  9  inches  ? 

Jins.  493  feet  2  inches  3  parts. 

4.  Required  the  area  of  a  four-sided  field,  whose 
south  side  is  2740  links,  east  side  3575  links,  north 


Mensuration  of  Superficies*  A3 

side  3755  links,  west  side  4105  links,  and  the*  diagonal 
from  south-west  to  north-east  4835  links. 

Jhis.  123  acres  0  rood  11.8673  perches. 

5.  Suppose  in  the  trapezium  ABCD  (see  the  fore- 
going figure)  the  side  AB  to  be  15,  BC  13,  CD  14, 
and  AD  12 ;  also  the  diagonal  AC  16  ;  what  is  the 
area  thereof?  Ms.  172.5247. 

6.  Suppose  in  the  trapezium  ABCD,  on  account  of 
obstacles,  I  could  only  measure  as  follows,  viz.  the 
diagonal  AC  878  yards,  the  side  AD  220  yards,  and 
the  side  BC  265  yards :  But  it  is  known  that  the  per- 
pendicular DE  will  fall  100  yards  from  A  ;  and  the 
perpendicular  BF  will  fall  70  yards  from  C  ;  required 
the  area  in  acres. 

Jlns.  The  perpendicular  DE  will  be  195.959  yards, 
BF  255.5875  yards  ;  and  the  area  of  the  trapezium 
85342.2885  yards,  or  17.6327  acres ;  or  17  acres,  2 
roods  21  perches. 

When  any  two  sides  of  a  Trapezium  are  parallel  to 
each  other,  it  is  then  generally  called  a  Trapezoid; 
the  area  may  be  found  by  the  following 

RULE. 

Multiply  half  the  sum  of  the  two  parallel  sides  by 
the  perpendicular  distance  between  them,  and  the  pro- 
duct will  be  the  area. 

EXAMPLES. 

1.  Let  x\3CD  be  a  trape- 
zoid, the  side  AB  23,  DC 
9.5,  and  CI  13,  what  is   the 


area 


First,  the  sum  of  AB  23, 
and  DC  9.5,  is  32.5 ;  the  half 
whereof  is  16.25,  which  mul- 
tipled  by  CI  13,  gives  the  product  211.25  for  the  area 
required. 


Mensuration  of  Superficies, 

Demonstration.  Bisect  AI)  in  II,  through  H  draw 
FE  parallel  to  BC,  ami  then  produce  CI)  to  E;  also 
draw  Gil  parallel  to  AB  or  CD.  Now  AH  is  equal 
to  HO.  and  the  angle  AUF  is  equal  to  the  ingle  EIIl) 
{Euclid.  I.  15.)  also  the  angle  HED  is  equal  to  the 
angle  UFA  {Euclid,!.  29.)  Therefore  the  triangles 
being  equiangular,  and  having  one  side  equal,  are 
equal  in  all  respects  {Euclid,!.  26.)  consequently  ED 
is  equal  to  AF.  Again,  IIG  is  half  the  sum  of  EC 
and  FB,  for  it  is  equal  to  each  of  them,  consequently 
it  is  half  the  sum  of  AB  and  DC,  but  FB  multiplied 
by  CI  is  the  area  of  the  figure  ECBF,  or  of  its  equal 
ADCB,  consequently  TIG  multiplied  into  CI  is  the 
area  thereof  5  which  was  to  be  proved. 

2.  Required  the  area  of  a  trapezoid  whose  parallel 
sides  are  750  and  1225  links,  and  the  perpendicular 
distance  between  them  1540  links  ? 

Jims,  15  acres  0  roods  33.2  perches. 

3.  What  is  the  Area  of  a  trapezoid  whose  parallel 
sides  are  12  feet  6  inches,  and  18  feet  4  inches ;  and 
the  perpendicular  distance  between  them  7  feet  9 
inches  ?  Jins.  119  feet  5  inches  9  parts. 

4.  A  field  in  the  form  of  a  trapezoid,  whose  parallel 
sides  are  6340  and  4380  yards,  and  the  perpendicular 
distance  between  them  121  yards,  lets  for  Z.S3  :  1  :  7^ 
fer  annum,  what  is  that  per  acre  ? 

Ms.  1.0  :  12  :  4f. 


§  VII.  Of  Irregular  Figures. 

Irregular  figures  are  all  such  as  have  more  sides 
than  four,  and  the  sides  and  angles  unequal.  All  such 
figures  may  be  divided  into  as  many  triangles  as  there 
are  sides,  wanting  two.  To  find  the  area  of  such 
figures,  they  must  be  divided  into  trapeziums  and  tri- 
angles, by  lines  drawn  from  one  angle  to  another ;  and 


Mensuration  of  Superficies.  S3 

so'find  the  areas  of  the  trapeziums  and  triangles  se- 
parately, and  then  add  all  the  areas  together;  so  will 
you  have  the  area  of  the  whole  figure. 

Let  ABCDEFG  be  an  irregular  figure  given  to  be 
measured  ;  first,  draw  the  lines  AC  and  GD,  and  there- 
by divide  the  given  figure  into  two  trapeziums,  A  CDG 
and  GDEF,  and  the  triangle  ABC;  Of  all  which,  I 
find  the  area  separately. 

First,  I  multiply  the  base  AC  by  half  the  perpen- 
dicular, and  the  product  is  49.6,  the  area  of  the  tri- 
angle ABC. 

Then  for  the  trapezium  ACDG,  the  two  perpen- 
diculars, 11  and  6.6,  added  together,  make  17.6 ; 
the  half  of  which  is  8.8,  multiplied  by  29,  the  dia- 
gonal ;  the  product  is  255.2,  the  area  of  that  trape- 
zium. 

And  for  the  trapezium  GDEF,  the  two  perpendicu- 
lars, 11.2  and  6,  added  together,  make  17.2;  the  half 
is  8.6;  which  multiplied  by  30.5,  the  diagonal,  the 
product  is  262.3,  the  area.  All  these  areas  added  to- 
gether, make  567.1,  and  so  much  is  the  area  of  the 
whole  irregular  figure.     See  the  work. 


06  Mensuration  of  Superficies. 

*>aso  AT.  11  perpendicular. 

-'  half  perpendicular.  o.r> 

49. G  area  of  ABC.  17.6  sum. 


8.8  half. 

29  diagonal  CG. 


792 
176 


£  perpendiculars. 


255.2  area  of  ACGD. 


11.2? v„.i„_  30.5 

8.6 


17.2  sum.  1830 

2440 

8.6  half  sum. 


262.30  area  of  GDEF. 
233.2     area  of  ACIMI. 
49.6        area  of  ABC. 


567.10  sum  of  the  areas. 

This  figure  being  composed  of  triangles  and  trape- 
ziums, and  those  figures  being  sufficiently  demonstrated 
in  the  Vth  and  Villi  sections  aforegoing,  it  will  be 
needless  to  mention  any  thing  of  the  demonstration  in 
this  place. 


§  VIII.  Of  Regular  Poltgqss. 

To  find  the  area,  or  superficial  content,  of  any  regu- 
lar polygon. 

RULE   i. 

Multiply  the  whole  perimeter,  or  sum  of  the  sides 
by  half  the  perpendicular  let  fall  from   the  centre 


•Mensuration  of  Superficies,  {,7 

to  the  middle  of  one  of  the  sides ;  and  the  product  is 
the  area. 


EXAMPLE.S 

1.  Let  HIKLMN  be  a  regular  hexagon,  each  side 
being  14.6,  the  sum  of  all  the  sides  is  87.6,  the  half 
sum  is  43.8,  which  mmltiplied  by  the  perpendicular 
GS  12.64397,  the  product  is  553.805886.  Or  if  87.6, 
the  whole  sum  of  the  sides,  be  multiplied  by  half  the 
perpendicular  6.321985,  the  product  is  553.805886, 
which  is  the  area  of  the  given  hexagon. 


By  Scale  and  Compasses. 

Extend  the  compasses  from  1  to  12.64,  that  extent 
will  reach  from  43.8,  the  same  way  to  553.8.  Or,  ex- 
tend from  %  to   12.64,  that  extent   will    reach  from 

K 


titration  of  Superficies. 

S7.r>  to  003.8.  Or,  as  before,  call  (he  first  number 
100  times  as  much  as  it  is,  and  each  of  the  two  follow- 
ing 10  times  as  much  as  they  are. 

Demonstration.  Every  regular  polygon  is  equal 
to  the  parallelogram,  or  Jong  square,  whose  length  is 
equal  to  half  the  sum  of  the  sides,  and  breadth  equal 
to  the  perpendicular  of  the  polygon,  as  appears  by  the 
foregoing  tigure  ;  for  the  hexagon  H1KLMN  is  made 
up  of  six  equilateral  triangles  :  and  the  parallelogram 
OPQU  is  also  composed  of  six  equilateral  triangles, 
that  is,  five  whole  ones,  and  two  halves  ;  therefore  the 
parallelogram  is  equal  to  the  hexagon. 

2.  The  side  of  a  regular  pentagon  is  25  yards,  and 
the  perpendicular  from  the  centre  to  the  middle  of  one 
of  the  sides  is  17.204775  ;  required  the  area. 

Jlns.  1075.2984. 

3.  Required  the  area  of  a  heptagon,  whose  side  is 
19.3S,  and  perpendicular  20.1215. 

'  Jlns.  136  l.S  11345. 

4.  Required  the  area  of  an  octagon,  whose  side  is 
9.941,  and  perpendicular  12.  Jlns.  477.168. 

5.  Required  the  area  of  a  decagon,  whose  side  is  20, 
and  perpendicular  30.776836.  Jlns.  3077.6836. 

6.  Required  the  area  of  a  duodecagon.  whose  side  is 
102,  and  perpendicular  190.3345908.  ' 

Jlns.  116484.7695696. 

7.  Required  the  area  of  a  nonagon,  whose  side  is 
40,  and  perpendicular  54.949548.       Jlns,  9890.91864. 


Mensuration  of  Superficies. 


99 


8.  Required  the  area  of  an  undecagon,  whose  side 
is  20,  and  perpendicular  34.056874. 

Ms.  3746.25614. 


9.  Required  the  area  of  a  trigon,  viz.  an  equilate- 
ral triangle,  whose  side  is  20,  and  perpendicular 
5.773502.  Ms.  173.20506, 


10.  Required  the  area  of  a  tetragon,  viz.  a  square, 
whose  side  is  20,  and  perpendicular  from  the  centre 
to  the  middle  of  one  of  the  sides  10*  Ms.  400. 


A  Table  for  the  more  ready  finding  the  Area  of  a  Po- 
lygon, and  also  the  Perpendicular. 


No.  of 

Name  of  the 

I.  Areas. 

II.  Perpend. 

Sides. 

Polygon. 

The  side  I. 

The  side  T. 

3 

Trigon, 

.433013 

.28867-51            ! 

4 

Tetragon, 

1.000000 

.5000000 

5 

Pentagon, 

1.720477 

.6881910 

6 

Hexagon, 

2.59S076 

.8660254           i 

7 

Heptagon, 

3.633912 

1.0382617           ! 

8 

Octagon, 

4.828427 

1.2071068 

9 

Nonagon, 

6.181824 

1.3737387 

10 

Decagon, 

7.694209 

1.5388418 

11 

Undecagon, 

9.365640 

1.7028437 

12 

Duodecagon, 

11.196152 

1.8660254          J 

RULE   II. 

Multiply  the  square  of  the  side  by  the  tabular  area, 
and  the  product  is  the  area  of  the  polygon.  Or,  mul- 
tiply the  side  of  the  polygon  by  the  tabular  perpendic- 
ular, and  the  product  will  give  the  perpendicular  of 
the  polygon  j  then  proceed  by  the  first  rule. 


1U0 


Mensuration  of  Superficies. 


How  to  find  the  Tabular  Numbers. 


These  numbers  are  found 
by  trigonometry,  thus  :  find 
the  ancle  at  the  centre  of 
the  polygon  by  dividing  360 
degrees  Dy  the  number  of 
sides  of  the  polygon. 


Example.  Suppose  each 
9ide  of  the  duodecagon  an- 
nexed be  l,  and  the  area  be 
required. 

Divide  360  by  12  (the  number  of  sides)  and  the  quo- 
tient is  30  degrees  for  the  angle  ACB ;  the  half  of 
which  is  15,  the  angle  DCB,  whose  complement  to  90 
degrees  is  75  degrees,  the  angle  CBD  :  then  say 


As  s,  DCB  15  degrees, 

is  to  .5  the  half-side  DB  log. 

so  is  s,  CBD  75  degrees, 


Co-ar. 


0.587004 
9.698970 
9.984944 


to  the  perpendicular  CD  1.866025        0.270918 

Then  1.866025  multiplied  by  6,  (the  half  perimeter) 
the  product  is  11.196152  the  area  of  the  duodecagon 
required. 

If  the  side  of  a  hexagon  be  14.6.  what 


Example. 
is  the  area  ? 
14.6 
14.6 

876 
584 
146 


213.16  squ. 


2.598076 
213.16 

15588456 
2598076 
7794228 
2398076 
5196152 


tab.  area. 


553.80588016  area,   the   same 
as  at  page   97. 


Mensuration  of  Superficies.  101 

Note.  Should  more  examples  he  thought  necessary, 
take  any  of  those  under  rule  I.     j    >  •*, 


§  IX.  Of  a  Circle. 


There  is  no  figure  that  affords  a  greater  variety  of 
useful  properties  than  the  circle.  It  is  the  most  ca- 
pacious of  all  plane  figures,  or  contains  the  greatest 
area  within  the  same  perimeter,  or  has  the  least  peri- 
meter about  the  same  area. 

The  area  of  a  circle  is  always  less  than  the  area  of 
any  regular  polygon  circumscribed  about  it,  and  its  cir- 
cumference always  less  than  the  circumference  of  the 
polygon.  But  on  the  other  hand,  its  area  is  always 
greater  than  that  of  its  inscribed  polygon,  and  its  cir- 
cumference greater  than  the  circumference  of  its  in- 
scribed polygon  ;  hence  the  circle  is  always  limited 
between  these  polygons.  The  area  of  a  circle  is  equal 
to  that  of  a  triangle,  whose  base  is  equal  to  the  cir- 
cumference, and  perpendicular  equal  to  the  radius. 

Circles,  like  other  similar  plane  figures,  are  in  pro- 
portion to  one  another,  as  the  squares  of  their  diame- 
ters ;  and  the  circumferences  of  circles,  are  to  one 
another  as  their  diameters,  or  radii. 

The  proportion  of  the  diameter  of  a  circle  to  its 
circumference  has  never  yet  been  exactly  determined. 
This  problem  has  engaged  the  attention,  and  exercised 
the  abilities  of  the  greatest  mathematicians  for  ages  ; 
no  square,  or  any  other  right-lined  figure,  has  yet 
been  found,  that  shall  he  perfectly  equal  to  a  given  cir- 
cle.   But  though  the  relation  between  the  diameter 

k2 


io J  Mensuration  of  Superficies, 

and  circumference  has  not  been  accurately  expressed 
itt. number?,  it  may  be  approximated  to  any  assigned 
.degree  of  exactilesV.  Jlrchimedes,  about  two  thousand 
the  proportion  to  be  nearly  as  7 
i  nearer,  ratios  have  since  been  suc- 
cessively assigned,  viz. 

As  106  to  333, 
As  113  to  355,  &C. 

This  last  proportion  is  very  useful,  for  being  turned 
into  a  decimal,  it  agrees  with  the  truth  to  the  sixth  fi- 
gure inclusively.  Victa,  in  his  universaliwn  inspectio- 
ncm  ad  canonem  mathematicum,  published  in  1579,  by 
means  of  the  inscribed  and  circumscribed  polygons  of 
363216  sides,  carried  the  ratio  to  ten  places  of  figures, 
shewing  that  if  the  diameter  of  a  circle  be  1  the  cir- 
cumference will 

be  greater  than  3.1415926535, 
but  less  than  3.1415926537. 

And  Ltidolf  Van  Ceulen,  in  his  book  de  circulo  et  ad- 
'scriptis,  by  the  same  means  carried  the  ratio  to  36  pla- 
ces of  figures ;  this  was  thought  so  extraordinary  a 
C performance,  that  the  numbers  were  cut  on  his  tomb- 
alone  in  St.  Peter's  church-yard,  at  Leyden.  These 
numbers  were  afterwards  confirmed  by  Willebrord 
Snell.  Mr.  Abraham  Sharp,  of  Little  Horton,  near 
Bradford,  Yorkshire,  extended  the  ratio  to  72  places 
of  figures,  by  means  of  Dr.  Halley's  series,  as  may  be 
seen  in  Sherwin's  logarithms.  Mr.  Machin,  professor 
of  astronomy  in  Gresham  college,  carried  the  ratio  to 
100  places  of  figures  ;  his  method  may  be  seen  in  Dr. 
Jlutton's  large  treatise  on  mensuration.  Lastly,  M.  de 
Lagny,  in  the  memoirs  de  VAcad.  1719,  by  means  of 
the  tangent  of  an  arch  of  30  degrees,  has  carried  the 


Mensuration  of  Superficies*  103 

ratio  to  the  amazing  extent  of  128  figures  ;  finding 
that  if  the  diameter  be  1,  the  circumference  will  be 
3.1415,  92653,  58979,  32384,  62643,  38327,  95028, 
84197,  16939,  93751,  05820,  97494,  45923,  07816, 
49628,    62089,    98628,    03482,  53421,    17067,  98214, 

80865,    13272,    30664,   70938,  446-f  or  447 But  on 

ordinary  occasions  3.1416  is  generally  used,  as  being 
sufficiently  exact. 

PROBLEM   I. 

Having  the  Diameter  to  find  the  Circumference,  or  cir- 
cumference to  find  the  diameter. 

RULES. 

1.  As  7  is  to  22,  so  is  the  diameter  to  the  circum- 
ference. 

Or,  as  113  is  to  355  so  is  the  diameter  to  the  cir- 
cumference. 

Or,  as  1  is  to  3.1416,  so  is  the  diameter  to  the  cir- 
cumference. 

2.  As  22  is  to  7,  so  is  the  circumference  to  the  di- 
ameter. 

Or,  as  355  is  to  113,  so  is  the  circumference  to  the 
diameter. 

Or,  as  3.1416  is  to  1,  so  is  the  circumference  to  the 
diameter ;  or,  which  is  the  same  thing,  as  1  is  to 
.318309,  so  i&  the  circumference  to  the  diameter. 

EXAMPLES. 


l.  The  diameter  of  a  circle  is  22.6?  what  is  the  cir- 
cumference ? 


Mensuration  of  Superf. 

Multiply  the  the  product  is 

:  which  divided  by  7  gives  71.02s  for  the  cii 

Or,  (by  the  second  proportion)  if  22.fi  he 
multiplied  by  835,  the  product  will  be  8023;  this  di- 
vided by  113,  the  quotient  is  71,  the  circumference. 
Or  (by  the  third  proportion)  if  22.6  be  multiplied  into 
3.1416,  the  product  is  71.00016,  the  circumference : 
the  two  last  proportions  are  the  most  exact. 


By  Scale  and  Compasses. 

Extend  the  compasses  from  7  to  22,  or  from  113  to 
355,  or  from  1  to  3.1416 ;  that  extent  will  reach  from 
22.6  to  71. 

2.  The  circumference  of  a  circle  is  71,  what  is  the 
diameter  ? 

If  .318309  he  multiplied  by  71,  the  product  will  be 
22.5999239  for  the  diameter.  Or,  113  multiplied  by 
71,  the  product  is  8023;  which  divided  by  355,  the 
quotient  will  be  22.6  the  diameter :  Or  7l  multiplied 
by  7,  the  product  is  497 ;  this  divided  by  22f  the  quo- 
tient as  22,5999,  the  diameter. 


Mensuration  of  Superficies.  10S 

By  Scale  and  Compasses, 

Extend  the  compasses  from  3.1416  to  1,  that  extent 
will  reach  from  71  to  22.6,  which  is  the  diameter 
sought. 

Or,  you  may  extend  from  1  to  .31S309. 

Or  from  22  to  7. 

Or  from  355  to  113  5  the  same  will  reach  from  71 
to  22.%  as  before. 

Note.  That  if  the  circumference  he  1,  the  diameter 
will  be  .318309 ;  this  number  being  found  by  dividing 
an  unit  by  3.1416. 

3.  If  the  diameter  of  the  earth  be  7970  miles,  what 
is  its  circumference,  supposing  it  a  perfect  sphere  ? 

Ms.  25038.552  miles. 

4.  Required  the  circumference  of  a  circle  whose 
diameter  is  50  feet.  Ms,  157.08  feet. 

5.  If  the  circumference  of  a  circle  be  12,  what  is 
tho  diameter  ?  Ms.  3.819708. 

PROBLEM   II. 

To  find  the  Area  of  a  Circle, 

RULES. 

1.  Multiply  half  the  circumference  by  half  the  dia- 
meter: Or,  take  £  of  the  product  of  the  whole  cir- 
cumference and  diameter,  and  it  will  give  the  area. 

2.  Multiply  the  square  of  the  diameter  by  '.7854,  and 
the  product  will  be  the  area. 

3.  As  452  is  to  355,  so  is  the  square  of  the  diameter 
to  the  area. 

4.  As  14  is  to  11,  so  is  the  square  ^>f  the  diameter 
to  the  area. 


106  Mensuration  of  Superf 

5.  Multiply  the   square   of    the    circumference   by 
58,  and  the  product  will  be  the  area. 

6.  As  88  is  to  7,  so  is  the  square  of  the  circumfer- 
ence to  the  area. 

7.  As  1420  is  to  113,  so  is  the  square  of  the  circum-  - 
Terence  to  the  area. 

EXAMPLES. 

1.  The  diameter  of  a  circle  is  22.6,  and  its  circum-  } 
ference  71,  what  is  the  area  by  the  first  rule  ? 


35.5  half  the 
11.3  half  diai 

circumference, 
meter. 

22.6  diameter. 
71  circumference. 

1065 
355 
355 

401.15  area. 

226 
1582 

4)1604.6 

401.15  area. 

Demonstration  of  rule  1.  Every  circle  may  be 
conceived  to  be  a  polygon  of  an  infinite  number  of 
sides  :  Now  the  semidiameter,  must  be  equal  to  the 
perpendicular  of  such  a  polygon ;  and  the  circum- 
ference of  the  circle  equal  to  the  periphery  of  the 
polygon;  therefore  half  the  circumference,  multiplied 
by  half  the  diameter,  gives  the  area.  The  other  part 
is  self-evident. 

2.  If  the  diaireter  of  a  circle  be  1,  and  its  circum- 
ference 3.1416,  what  is  the  area  ? 

Jins,  .7854. 


Mensuration  of  Superficies.  107 

3.  If  the  diameter  of  a  circle  be  22.6,  what  is  the 
area  by  the  second  rule  ? 


22.6  diameter.  510.76 

22.6  .7854 


1356  204304 

452  255380 

452  40860S 


357532 


510.76  squ.  of  the  diameter. 


401.150904  area. 

\ 

• 
Demonstration  of  rule  2.  All  circles  are  to  each 
other  as  the  squares  of  the  diameters,  and  the  area  of 
a  circle  whose  diameter  is  1,  is  .7854  (by  the  second 
example);  therefore  as  the  square  of  1,  which  is  1, 
is  to  .7854,  so  is  the  square  of  the  diameter  of  any 
circle  to  its  area. 

4.  If  the  diameter  of  a  circle  be  22.6,  what  is  the 
,area  by  the  third  rule  ? 

As  452  :  355  ::  510.76  the  square  of  the  diameter: 
401.15,  the  same  as  before. 

Demonstration  of  rule  3.  By  problem  I.  we  have, 
as  113  :  355  ::  diameter  1  :  circumference;  but  the 
area  of  a  circle  whose  diameter  is  1,  is  *  of  this  cir- 
cumference by  rule  1.  problem  II.  viz.  it  is  ||| ;  but 
the  areas  of  circles  are  to  each  other  as  the  squares 
of  their  diameters,  therefore  square  1:  |f|:  :  square 
of  any  diameter  :  the  area ;  viz.  452  :  355 :  :  square 
of  any  diameter  :  the  area. 

5.  If  the  diameter  of  a  circle  be  22.6,  what  is  the 
area  by  the  fourth  rule  ? 


108  Mensuration  of  Superjl 

As  14  :  11  ::  610.7ft  the  square  of  the  diameter  : 
401.31  the  area  which  is  larger  than  by  the  other  me- 
thods. 

Demonstration  of  rule  4.  By  problem  I.  we  have 
as  7  :  23  ::  diameter  1  :  circumference ;  and,  by  the 
same  argument  as  above,  the  area  of  a  circle  whose 
diameter  is  1,  is  ||  or  \l»  hence,  14:  11 ::  square  of 
the  diameter  :  the  area. 

6.  If  the  circumference  of  a  circle  be  71,  what  is  the 
area  by  rule  5  ? 

71  circumference.  .07958 

71  0041 


71  7958 

497  31833 


5041  squ.of  the  circumference. 


39790 


401.16278  area. 


The  reason  of  the  above  operation  is  easy ;  for  as 
the  diameter  of  one  circle  is  to  its  circumference,  so 
is  the  diameter  of  any  other  circle  to  its  circumfer- 
ence :  therefore  the  areas  of  circles  are  to  each  other 
as  the  squares  of  their  circumferences ;  and  .07958  is 
the  area  of  a  circle  whose  circumference  is  1 ;  for 
when  the  circumference  is  1,  the  diameter  is  .318309, 
as  has  been  observed  before. 

7.  If  the  circumference  of  a  circle  be  71,  what  is 
the  area  by  the  sixth  rule  ? 

As  ss  :  7  ::  5041  the  square  of  the  circumference : 
400.98  area. 


Mensuration  of  Superficies.  109 

Demonstration  of  rule  6.  By  problem  I.  we 
have  as  22  :  7  ::  circumference  1  :  diameter  ;  but  by 
rule  I.  problem  II.  the  area  of  a  circle  whose  circum- 
ference is  1,  is  i  of  this  diameter,  viz.  ¥\.  Now  the 
areas  of  circles  are  to  each  other  as  the  squares  of 
their  circumferences;  therefore  square  1  :  -^  ::  square 
of  any  circumference:  area;  viz.  88:  7  ::  square  of 
any  circumference  :  area.  vSk,* 

8.  If  the  circumference  of  a  circle  be  71,  wh^t  is  the 
area  by  the  7th  rule  ?  ** 

As  1420  :  113  ::  5041  the  square  of  the  circumfer- 
ence :  401.15  area, 

Demonstration  of  rule  7.  By  problem  I.  we  have 
as  355  :  113  ::  circumference  1  :  diameter;  and,  by  the 
same  argument  as  above,  the  area  of  a  circle  whose 
circumference  is  1,  is  tVt3o  ?  therefore  it  follows,  that 
1420  :  113  ::  square  of  any  circumference :  area. 

9.  What  is  the  area  of  a  circle  whose  diameter  is 
12?  Ms.  113.0976. 

10.  The  diameter  of  a  circle  is  12,  and  its  circum- 
ference 37.6992,  what  is  the  area  ? 

Jlns.  113.0976. 

11.  The  circumference  of  a  circle  is  37.6992,  what 
is  the  area?  Jlns.  113.104579,  &c. 

12.  The  surveying-wheel  is  so  contrived  as  to  turn 
just  twice  in  length  of  a  pole,  or  16 £  feet;  in  going 
rou^d  a  circular  bowling-green,  I  observed  it  to  turn 
exactly  200  times;  what  is  the  area  of  the  bowling- 
green  ? 

Ans.  4  acres  3  roods  35.8  perches. 

13.  The  length  of  a  line,  with  which  my  gardener 
formed  a  circular  fish  pond,  was  exactly  27|  yards : 
pray  what  quantity  of  ground  did  the  fish-pond  take 
,iin  ? 

Jlns.  2419.22895  yards,  or  half  an  acre  nearly. 


no  Mensuration  of  Superficies. 


PROBLEM    III. 


Having  the  Diameter,  Circumference,  or  Jirea  of  a  Cir- 
cle given  :  to  find  the  side  of  a  Square  equal  in  Jlrea 
to  the  Circle,  and  the  side  of  a  Square  inscribed  in 
the  Circle  ;  Or,  having  the  side  of  a  Square  given, 
to  find  the  Diameter  of  its  Circumscribing  Circle, 
and  also  of  a  Circle  equal  in  Area,  <$*c. 

RULES. 

1.  The  diameter  of  any  circle,  multiplied  by 
.8862269,  will  give  the  side  of  a  square  equal  in 
area. 

2.  The  circumference  of  any  circle,  multiplied 
by  .2820948,  will  give  the  side  of  a  square  equal  in 
area. 

3.  The  diameter  of  any  circle,  multiplied  by 
.707106S,  will  give  the  side  of  a  square  inscribed  in 
that  circle. 

4.  The  circumference  of  any  circle,  multiplied  by 
.2250791,  will  give  the  side  of  a  square  inscribed  in 
that  circle. 

5.  The  area  of  any  circle,  multiplied  by  .6366197, 
and  the  square  root  of  the  product  extracted,  will  give 
the  side  of  a  square  inscribed  in  that  circle. 

6.  The  side  of  any  square,  multiplied  by  1.414236, 
will  give  the  diameter  of  its  circumscribing  circle. 

7.  The  side  of  any  square  multiplied  by  4.4428829, 
will  give  the  circumference  of  its  circumscribing 
circle. 

8.  The  side  of  any  square,  multiplied  by  1.1283791, 
will  give  the  diameter  of  a  circle  equal  in  area  to  the 
square. 

9.  The  side  of  any  square,  multiplied  by  3.5449076, 
will  give  the  circumference  of  a  circle  equal  in  areato 
the  square. 


Mensuration  of  Superficies,  111 

Explanation  of  these  Rules. 

Let  ABBC  represent  a  square  equal  in  area  to  the 
circle;  and  EH.FG  a  square  inscribed  in  the  circle ; 
GH  and  EF  diameters  of  the  circle,  perpendicular  to 
eaeh  other. 

1.  The  area  of  a  circle,  whose  diameter  is  1,  is 
►7854,  or  .78539816,  &c.   the   square   root  of  which 

gives  the  number  in  rule  1 And  as  the  diameter  of 

one  circle  is  to  the  diameter  of  another,  so  is  the  side 
of  a  square  equal  in  area  to  the  former,  to  the  side  of 
a  square  equal  in  area  to  the  latter,  &c. 


2.  The  area  of  a  circle,  whose  circumference  is  1 
is  .07  95775,  &c.  the  square  root  of  which  gives  the 
number  in  the  2d  rule. 

3.  To  the  square  of  EO=|,  add  the  square  of 
GO=|,  the  square  root  of  the  sum  will  give  the  third 
number. 

4.  The  diameter  of  any  circle,  when  the  circum- 
ference is  1,  is  .318309,  &e.  hence  EO  or  GO  is 
1.159154,  &c.  with  which  proceed  as  above,  and  you 
will  iind  the  fourth  number. 


•J  13  Mensuration  of  Superficies. 

Hence  it  appears,  that  if  the  diameter  of  the  cir- 
cle is  given,  or  can  be  found,  the  side  of  the  inscri- 
bed square  may  be  found  ;  and  if  (he  side  of  a  square 
equal  in  area  to  the  circle  is  given,  the  diameter  and 
circumference  of  the  circle  may  be  readily  found,  the 
former  by  dividing  the  area  by  .78539,  &c.and  extract- 
ing the  square  root  of  the  quotient  (see  rule  2.  prob.  II.) 
and  the  latter  by  dividing  the  area  by  .0795775,  &c. 
and  extracting  the  square  root  of  the  quotient  (see 
rule  5.  prob.  II.) — The  rest  of  the  numbers  are  left 
for  the  learner  to  examine  at  his  leisure. 


EXAMPLES. 

4.  If  the  diameter  of  a  circle  be  22.6,  what  is  the 
side  of  a  square  equal  in  area  to  the  circle  ? 

The  side  of  a  square  equal  in  area  to  a  circle,  whose 
diameter  is  1,  by  rule  1,  is  .8862269; — hence,  22.6X 
8862269=20.02872794,  the  side  >f  the  square. 

2.  If  the  diameter  ef  a  circle  be  50,  what  is  the 
side  of  a  square  equal  in  area  ?  Ms.  44.311345. 

3.  If  the  circumference  of  a  circle  be  40,  what  is 
the  side  of  a  square  equal  in  area  to  the  circle  ? 

The  side  of  a  square  equal  in  area  to  a  circle  whose 
circumference  is  1,  by  rule  2,  is  .2820948;  and 
.2820948X40=11.283792,  the  side  of  the  square. 

4.  If  the  circumference  of  a  circle  be  71,  what  is 
the  side  of  a  square  equal  in  area  to  the  circle  ? 

Ms.  20.0287308. 

5.  If  the  diameter  of  a  circle  be  50,  what  is  the  side 
of  a  square  inscribed  in  that  circle  ? 

The  side  of  a  square  inscribed  in  a  circle  whose 
diameter  is  l,  by  rule  3,  is  .7071068,  therefore,  50 x 
.7071068=35.35534,  the  side  of  the  square. 

6.  If  the  diameter  of  a  circle  be  22.6,  what  is  the 
side  of  a  square  inscribed  in  that  circle  ? 

Ms.  15.98061368. 


Mensuration  of  Superficies.  118 

7.  If  the  circumference  of  a  circle  be  40,  what  is 
the  side  of  a  square  inscribed  in  that  circle  ? 

The  side  of  a  square  inscribed  in  a  circle  whose 
circumference  is  1,  by  rule  4,  is  .2250791  ;  and  40X 
.22507011=9,003164,  the  side  required. 

8.  If  the  circumference  of  a  circle  be  71,  what  is 
the  side  of  a  square  inscribed  in  that  circle  ? 

Ms.  15.9806161. 

9.  If  the  area  of  a  circle  be  1963.5,  what  is  the 
side  of  a  square  inscribed  in  that  circle  ? 

The  area  of  a  square  inscribed  in  a  circle  whose 
area  is  1,  by  rule  5,  is  .6366197;  therefore,  1963. 5X 
.6366197=1250.00278095,  whose  square  root  is 
35.3553,  the  side  of  the  inscribed  square. 

10.  If  the  area  of  a  circle  be  401.15,  what  is  the 
side  of  a  square  inscribed  in  that  circle. 

Jns.  15.9806. 

11.  If  the  side  of  a  square  be  35.36,  what  is  fhe 
diameter  of  a  circle  which  will  circumscribe  that 
square  ? 

The  diameter  of  a  circle  that  will  circumscribe  a 
square  whose  side  is  1,  by  rule  6,  is  1.4142136 ;  and 
.35.36X1-4142136=50.006592896,  the  diameter. 

12.  If  the  side  of  a  square  be  15.98,  what  is  the 
diameter  of  a  circle  which  will  circumscribe  that 
square.  Ans.  22.599133328. 

13.  If  the  side  of  a  square  be  35.36,  what  is  the 
circumference  of  a  circle  which  will  circumscribe  that 
square  ? 

The  circumference  of  a  circle  that  will  circum- 
scribe a  square  whose  side  is  1,  by  rule  7,  is  4.4428829 
and  35.36X4-4428829x^157.1000339344,  the  circum- 
ference. 

14.  If  the  side  of  a  square  be  15.98,  what  is  the 
circumference  of  a  circle  which  will  circumscribe 
that  square  ?  Ans.  70.997268742. 

15.  If  the  side  of  a  square  be  35.36,  what  will  be 
the  diameter  of  a  circle  having  the  same  area  as  the 
square  ? 

l  2 


114  Mensuration  of  Superficies. 

The  diameter  of  a  circle  having  the  same  area  as 
a  square  whose  side  is  1,  by  rule  8,  is  1.1283791; 
hence  35.36X1.1 -83791=39.899484976,  the  diameter. 

IK.  It'  the  side  of  a  square  be  20.029,  what  will  be 
the  diameter  of  a  circle  having  the  same  area  as  the 
square  ?  Ms.  22.6003049939. 

17.  If  the  side  of  a  square  he  33.36,  what  is  the 
circumference  of  a  circle,  having  the  same  area  as  the 
square  ? 

The  circumference  of  a  circle  that  will  be  equal  to 
the  area  of  a  square  whose  side  is  1,  by  rule  9,  is 
3.5449076  ;  therefore — 

35.36X3.5449076=  125.347932736,  the  side 
of  the  square. 

18.  If  the  side  of  a  square  be  20.029,  what  is  tli3 
circumference  of  a  circle,  having  the  same  area  as  the 
square  ?  Jins.  71.000954. 

Note.  All  the  foregoing  examples  are  easily  work- 
ed by  scale  and  compasses,  for  they  consist  only  of  mul- 
tiplication. But  in  the  9th  example  a  square  root  is 
to  be  extracted ;  and  in  order  to  do  this  right,  it  will 
he  necessary  to  call  the  middle  unit  upon  the  scale  a 
square  number,  as  1,  100,  10000,  &c.  or  1,  T^T,  &c — 
Thus,  if  the  number  to  be  extracted  consist  of  one 
whole  number,  call  the  1  in  the  middle  of  the  scale  an 
unit;  if  of  two  or  three  whole  numbers,  call  it  100; 
if  four  or  five  whole  numbers,  call  it  10000,  &c.  for 
whole  or  mixed  numbers.  For  pure  decimals,  if  the 
number  to  be  extracted  consist  of  one  or  two  places, 
call  the  unit  in  the  middle  of  the  scale  1,  then  the  1 
at  the  left-hand  will  be  T\j,  the  figure  2  will  be  T90,  &c. 
or  the  l  at  the  left-hand  may  be  called  T^,  then  the 
figure  2  will  be  T2^.  &c.  so  that  you  will  more  readily 
obtain  the  intermediate  places. — This  being  under* 
stood,  extend  the  compasses  from  the  middle  unit  thus 
estimated,  to  the  number  to  be  extracted  ;  divide  the 
space  of  this  extent  into  two  equal  parts,  and  the  m&V 


Mensuration  of  Superficies.  115 

die  point  will  be  the  root  required  :  but  here  you 
must  observe  that  the  middle  unit  will  change  its  val- 
ue (except  you  call  it  an  unit),  that  is,  if  in  the  first 
extent  you  estimated  it  at  100,  it  must  now  be  consid- 
ered as  10;  if  at  10000,  it  must  now  be  called  100,  &c. 
Thus  in  the  9th  example  the  square  root  of  1250  is  to 
he  extracted ;  here  I  consider  the  middle  unit  as  10000, 
and  extend  from  it  towards  the  left-hand  to  1250,  the 
middle  point  between  these  is  35.3,  estimating  the  mid- 
dle unit  now  at  100.  A  little  practice  will  render 
these  observations  familiar. 


§  X.  To  find  the  Area  of  a  Semicircle. 


Multiply  the  fourth  part  of  the  circumference  of 
the  whole  (that  is,  half  the  arch  line)  by  the  seinidi- 
ameter,  the  product  is  the  area. 

EXAMPLES. 

1.  Let  ACB  be  a  semi- 
circle, whose   diameter   is  ^    *L 
22.6,  and  the  half  circum- 
ference, or  arch  line,  ACB, 
is    35.5,    the   half  of  it  is 
17.75,    which  multiply   by       /  iM 
the  semidiameter  ll.a,  and      y^amo^J^. 
the  product  is  200.575,  the  area  of  the  semicircle. 


By  Scale  and  Compasses.        N 

Extend  the  compasses  from  1  to  11.3;  that  extent 
will  reach  from  17.75  to  200.575,  the  area.  Or,  as 
before,  call  the  first  number  100  times  as  much  as  it  is> 


116  Mensuration  of  Superficies. 

and  each  of  the  two  following  10  times  as  much  as 
they  are. 

If  only  the  diameter  of  the  semicircle  he  given 
you  may  say,  by  the  Rule  of  Three, 

As  1  is  to  .39  27,  so  is  the  square  of  the  diameter  to 
the  area ;  or  multiply  the  square  of  the  diameter  by 
.7854,  and  take  half  the  product. 

2.  The  diameter  of  a  circle  is  30,  and  the  semicir- 
cumference  47.124 ;  what  is  the  area  of  the  semicir- 
cle ?  Ms.  353.43. 

3.  The  diameter  of  a  circle  is  200,  what  is  the  area 
of  the  semicircle  ?  Jlns.  15708. 

4.  If  the  circumference  of  a  circle  he  157.08,  and 
the  diameter  50 ;  what  is  the  area  of  the  semicircle  ? 

dns.  981.75. 


§  XI.  To  find  the  Area  of  a  Quadrant. 
rule. 

Multiply  half  the  arch  line  of  the  quadrant,  (that 
is,  the  eighth  part  of  the  circumference  of  the  whole 
circle,)  by  the  semidiameter,  and  the  product  is  the 
area  of  the  quadrant. 

EXAMPLES. 

1.  Let   ABC  be  a  quadrant,   or     _ 
fourth  part  of  a  cirele,  whose  radi- 
us, or  semidiameter  is  11.3,  and  the 
half  arch  line  8.875;  these  multipli- 
ed together,  the  product  is  100.2875  \ 
for  the  area.  \ 

These  are  the  rules  commonly  given  for  finding  the 


Mensuration  of  Superficies.  117 

area  of  a  semicircle  and  quadrant;  but  it  is  the  best 
way  to  find  the  area  of  the  whole  circle,  and  then 
take  half  that  area  for  the  semicircle,  and  a  fourth 
part  for  the  quadrant. 

2.  Required  the  area  of  a  quadrant,  the  radius  being 
15,  and  the  length  of  the  arch  line  23.562. 

Ms.  176.710. 

3.  The  diameter  of  a  circle  is  200,  what  is  the  area 
of  the  quadrant  ?  Ms.  7854. 

4.  If  the   circumference  of  a  circle   be  157.08,  and 
the  diameter  50,  what  is  the  area  of  the  quadrant  ? 

Ms.  490.875. 
Before  I  proceed  to  shew  how  to   find  the  area  of 
the  sector,  and  segment  of  a  circle,  I  shall  shew  how 
to  find  the  length  of  any  arch  of  a  circle. 


To  find  the.  length  of  any  Mch  of  a  Cirde. 


RULE    I. 

Multiply,  continually,  the  radius,  the  number  of  de- 
grees in  the  given  arch,  and  the  number  *.0l745329  ; 
the  product  will  be  the  length  of  the  arch. 

Example.  1.  If  the  arch  ACB  contain  118  degrees 
46  minutes  40  seconds,  and  the  radius  of  the  circle,  of 
which  ACB  is  an  arch,  be  19.9855:  what  is  the  length 
of  the  arch  ? 

First,  118  degrees  46  minutes  40  seconds,  are  equal 
to  118|  degrees. 

19.9855X118|X-0l745329=r41.431199,     &C.    the 
length  of  the  arch. 


*  When  the  radms  is  1,  the  semi-circumference  of  the  cir- 
cle is  3.14159265,  &c.  and  this  number  divided  by  180,  the 
degrees  in  a  semicircle,  gives  .01745329  for  the  length  of  one 
degree  when  the  radius  is  unity. 


113 


Mensuration  of  Superficies, 


RULE    II. 

From  eii^lit  times  the  chord  of  half  the  arch,  sub- 
tract t  he  chord  of  the  whole  arch,  ami  one  third  of  the 
remainder  will  be  the  length  of  the  arch  nearly. 

Example  2.  If  the  chord,  AC  or  BO,  of  half  the 
arch  be  1  .8,  and  the  chord  AB  of  the  whole  arch  34,4  j 
what  is  the  length  of  the  arch  ACB  ? 


8X19.8 — 34.4  —  124  j  which  divided  by  5,  gives  41  j- 
for  the  length  of  the  arch. 

3.  If  the  chord  AB  of  the  whole  arch  be  6,  and  the 
chord  AC  of  half  the  arch  3.0*3836,  what  is  the  length 
of  the  arch  ?  Ms.  6.116896. 

4.  If  the  radius  of  the  circle  be  9,  and  the  arch 
contain  38.9424412  degrees,  what  is  its  length  ? 

Ms.  6.117063. 

5.  There  is  an  arch  of  a  circle  whose  chord  AB  is 
50.8  inches,  and  the  chord  AC  30.6,  what  is  the  length 
of  the  arch  ?  Ms.  64|  inches. 

6.  The  chord  AB  of  the  whole  arch  is  40,  and  the 
versed  sine  DC  15,  what  is  the  length  of  the  arch  ? 


•ins. 


5B±. 


7.  If  the  radius  of  the  circle  be  11.3,  and  the  arch 
contain  52  degrees  15  minutes,  what  is  its  length  ? 

Ms.  10.304858,  &C. 


Mensuration  of  Superficies. 


119 


To  find  the  diameter  of  a  Circle,  by  having  the  chord  and 
versed  sine  of  the  Segment  given. 


RULE. 


Square  half  the  chord ;  divide  it  by  the  versed  sine, 
and  to  the  quotient  add  the  versed  sine  ;  the' sum  will 
be  the  diameter. 


Demonstration. 
The  half  chord  AD, 
is  a  mean  propor- 
tional between  the 
segments,  CD,DE, 
of  the  diameter  CE; 
that  is  AD*=CD 
XDE  (  by  13.6.  ) 

AD* 
therefore,     = 

CD 
DE ;  to  these  equal 
AD 

add  CD,  and 

CD 
xCD=DExDC 

:CE  :  which,  in  words,  gives  the  above  rule. 


'      EXAMPLES. 

1.  Let  ACB  be  a  segment  given,  whose  chord  AB 
is  36,  and  the  versed  sine  CD  6;  half  36  is  18,  which 
squared,  makes  324;  this  divided  by  6,  the  quotient  is 
04 :  to  which  add  6,  the  sum  is  60,  the  diameter  of 
the  circle  CE. 

2.  If  the  chord  AB  be  18,  and  versed  sine  DC  4  ; 
what  is  the  diameter  EC  ?  dns.  2±\, 


120  Mensuration  of  Stiperjl 

§  XII.     To  find  the  Area  of  the  Sector  of  a  Circle. 

rule. 

Multiply  lialf  the  length  of  the  arch  by  the  radius 
of  the  circle,  and  the  product  is  the  area  of  the  sector. 

EXAMPLL3. 

I.  Let  ADBE  he  the  sector  of  a  circle  given,  whose' 
radius  AE  or  BE  is  24.5,   and  the  chord  AB  39.2  ; 
what  is  the  area  ? 

First,  find  the  chord  of  half  the  arch,  and  then  the 
length  of  the  arch  by  section  XI. 

From  the  square  of  AE  600.2.5,  take  the  square  of 
AC  38*.  16,  the  square  root  of  the  remainder,  216.09. 
gives  EC,  14.7  ;  from  ED,  or  its  equal  AE,  take  EC, 
the  remainder  9.8  is  CD — then  to  the  square  of  AC 
384.16,  add  the  square  of  CD  96.04,  the  square  root 
of  the  sum  480.2  gives  AD  21.9135,  the  chord  of  half 
the  arch,  hence  the  arch  AD  is 
found  to  be  22.68466  ;  this  multi- 
plied by  the  radius,  24.5,  gives 
555.77417  for  the  area  of  the  sec- 
tor. , 


2.  LetLMNO  be  a  sector  greater  than  a  semicircle,^ 
whose  radius  LO  or  NO,  is  20.6,  and  the  chord  Nc  of 
half  the  arch  McN  20.35962;  also  the  chord   MN  of 
the  whole  arch  McN,  that  is,  of  half  the  arch  of  the 
sector,  is  33.4;  what  is  the  area? 


Mensuration  of  Superficies. 


12i 


The  arch  McL  will  be  found,  by  the  following,  to 
be  42.49232,  this  multiplied  by  the  radius,  20.6,  gives 
875.341792,  the  area. 


3.  Required  the  area  of  a  sector,  less  than  a  se- 
micircle, whose  radius  is  9,  and  the  chord  of  the 
arch  6. 

Ms.  The  length  of  the  arch  is  6.116896,  and  area 
27.526032. 

4.  Required  the  area  of  a  sector,  whose  arch  con- 
tains 18  degrees,  the  radius  being  3  feet. 

Jlns.  The  length  of  the  arch  is  .£4247766,  and  the 
area  1.41371649. 

5.  Required  the  area  of  a  sector,  greater  than  a 
semicircle,  whose  radius  LO  or  NO  is  10,  and  the 
chord  MN  of  half  the  arch  LM  c  N,  viz.  the  chord  of 
the  whole  arch  M  c  N,  is  16. 

Jlns.  As  in  example  1,  you  will  find  Nc= 8.9442719, 
and  the  length  of  the  arch  McN  18.518050384; 
hence  the  whole  area  will  be  185.18050584. 

6.  What  is  the  area  of  a  sector,  less  than  a  semi- 
circle, the  chord  of  the  whole  arch  AB  being  50.8, 
and  the  chord  of  half  the  arch  AD  30.6. 

Jlns.  The  length  of  the  arch  is  64§  \  and  by  taking 
the  square  root  of  the  difference  of  the  squares  of  AD 

M 


I 


1^2  Jlensurution  of  Supcrjl 

and  AC,  DC  will  be  found  to  be  t7.or>i5;  then  by  the 
method  of  finding  a  diameter,  Section  XI.  as  17.0940: 
35.4  (=AC)  ::  20.4  :  37.8071,  the  remaining  part  of 
the  diameter;  hence  the  diameter  is  5  4.8716,  radius 
27A.i3S,  and  area  of  the  sector  887.0908. 


§XIII.  To  find  the  Area  of  the  Segment  of  a 
Circle. 

rule  I. 

Find  the  area  of  the  whole  sector  CADBC  by 
Section  XII.  and  then  (by  Section  V.)  find  the  area  of 
the  triangle  ABC,  and  subtract  the  area  of  the  trian- 
gle from  the  area  of  the  sector,  the  remainder  will  be 
the  area  of  the  segment. — If  the  segment  be  greater 
than  a  semicircle,  add  the  area  of  the  triangle  to  the 
area  of  the  sector,  and  the  sum  will  be  the  area  of 
the  segment. 

EXAMPLES. 

1.  Required  the  area 
of  the  segment  AD  BE 
whose  chord  AB  is  35, 
and  versed  sine  DE,  9.6. 

First.  By  the  method  of  ^ 
finding  a  diameter,  sect. 
XI.  as  9.6:  17.5  (=AE) 
::  17.5  :  31.9,  the  re- 
maining part  of  the  dia- 
meter, which,  added  to 

DE,  gives  41.5  for  the  diameter,  the  half  of  which, 
20.75,  is  the  radius  CD ;  from  which  take  DE,  the 
remainder  11.15,  is  the  perpendicular  CE  of  the  tri- 
angle ACB. 


Mensuration  of  Superficies. 


123 


Second.  To  the  square  of  AE  306.23  add  the  square 
of  ED  92.16,  the  square  root  of  the  sum  398.41  is  19.96, 
the  chord  AD  of  half  the  arch :  then,  the  area  of  the 
triangle  ABC  will  be  =  195.125  the  arch  AD=20.7S; 
the  area  of  the  sector  CADB=43 1.1850$  and  that  of 
the  segment  ABD,= 236.06. 

2.  Let  MACBM  be  a  segment  greater  than  a 
semicircle ;  there  are  given  the  base  of  the  segment 
or  chord  AB  20.5,  the  height  MC  17.17,  the  radius 
of  the  circle  1 1 .64,  the  chord  of  half  the  arch  ACB, 
viz.  AC  20,  and  the  chord  of  one -fourth  of  the  areh 
11.5,  to  find  the  area  of  the  segment. 


Half  the  arch  line 
will  be  found,  as  before, 
to  be =24  ;  which  mul- 
tiplied by  the  radius 
1 1.64,  gives  279.36  for 
the  area  of  the  sector 
LACB;  and  half  the 
base  AM,  multiplied  by 
the  perpendicular  LM? 
=  5.53  gives  56.6825  for 
triangle 


the  area  of  the 

ABL.    Hence  336.0425  is  the 

ABC. 


area  of  the  segment 


RULE   II. 

To  two  thirds  of  the  product  of  the  chord  and 
height  of  the  segment,  add  the  cube  of  the  height 
divided  by  twice  the  chord,  and  the  sum  will  be  the 
area,  nearly. 

Note.  When  the  segment  is  greater  than  a  semicir- 
cle, find  the  area  of  the  remaining  segment,  and  sub 
tract  it  from  the  area  of  the  whole  circle. 


fgft  Mensuration  of  Sii])erf< 

EXAMPLES. 

1.  Required  the  area  of  a  segment  ADBE,  less  than 
a  semicircle,  whose  chord  AB  is  3D,  and  versed  sine, 
or  height  of  the  segment,  DE  9.6. 

|  of  35  X  9.6=r224,  and 
9.6X  9.6  X  9.6—2  X  35  =  12.639 


Area  of  the  segment=236.639 

2.  The  area  of  a  segment  greater  than  a  semi- 
circle is  required,  the  chord  being  20.5,  and  height 
17.17. 

As  17.17:  10.25  ::  10.25  :  6.119,  the  height  of  the 
remaining  segment  by  sect.  XI ;  hence  the  diameter  is 
23.289,  and  the  area  of  the  whole  circle  425.983305. 
By  rule  2,  the  area  of  the  remaining  segment  is 
89.211347,  which  subtracted  from  the  area  of  the 
whole  circle,  leaves  336.768958  for  the  area  required- 
See  example  2. 

3.  Required  the  area  of  a  segment,  less  than  a  semi- 
circle, whose  chord  is  18.9,  and  height  2.4. 

„        530.601875  by  rule  1. 
**"S'    I  30.6057  by  rule  2. 

4.  Required  the  area  of  a  segment,  less  than  a  se- 
micircle, whose  chord  is  23,  and  height  3.5. 

a        5  54.5844  by  rule  1. 
"n      I  54.598732  by  rule  2. 

5.  Required  the  area  of  a  segment,  greater  than  a 
semicircle,  whose  chord  is  12,  and  height  18. 

Jlns.  297.826  by  rule  2.  This  may  also  be  done  by 
rule  1 ;  for,  by  what  is  given,  every  given  line  marked 
in  the  figure  to  example  2,  may  be  found. 


Mensuration  of  Superficies.  123 


§XIV.  To  find  the  Area  of  Compound  ^Figures. 

Mixed  or  compound  figures  are  such  as  are  compos- 
ed of  rectilineal  and  curvilineal  figures  together. 

To  find  the  area  of  such  mixed  figures,  you  must 
find  the  area  of  the  several  figures  of  which  the  whole 
compound  figure, is  composed,  and  add  all  the  areas 
together,  and  the  sum  will  be  the  area  of  the  whole 
compound  figure. 

Let  AaBCcDA  he  a  compound  figure,  AaB  being 
the  arch  of  a  circle  whose  chord  AB  is  18.9,  and 
height  2.4  ;  CcD  is  likewise  the  arch  of  a  circle  whose 
chord  CD  is  23,  and  height  3.5 :  And  ABCD  is  a  tra- 
pezium, whose  diagonal  BD  is  34,  and  the  two  per- 
pendiculars 16  and  16.2  3  required  the  area  of  this 
compound  figure. 


By  example  4.  section  XIII.  the  area  of 
the  segment  A«B  is 30.60187.5 

By  example  5.  section  XIII.  the  area  of 
the  segment  CcD  is 54.5844 

Area  of  the  trapezium  ABCD        -     -      547.4 


Area  of  the  compound  figure    -    -    -    632.586275 

M  2 


1 26  Mensuration  of  Superficies. 

> 

§  XV.  To  find  the  Area  of  an  Ellipsis. 

rule. 

Multiply  the  transverse,  or  longer  diameter,  hy  the 
conjugate  or  shorter  diameter,  and  that  product  by 
7851 5  the  last  product  is  the  area  of  the  ellipsis. 


&0 


5 


EXAMPLES. 

1.  Required  the  area  of  an   ellipsis  whose  longer 
diameter  AB  is  61.6,  and  shorter  diameter  CD  44.4. 

61.6  X44.4X  .7854=  2148.100416,   the   area   of  the 
ellipsis. 

2.  The  longer  diameter  of  an  ellipsis  is  50,  and  the 
shorter  40,  what  is  the  area?  Am.  1570.8. 

3.  The  longer  diameter  of  an  ellipsis  is  70,  and  the 
shorter  50,  what  is  the  area  ?  Am.  2748.9. 

4.  Required  the  area  of  an  ellipsis,  whose  transverse 
diameter  is  24,  and  conjugate  18. 

Ans.  339.2928, 


Mensuration  of  Superficies. 
Demonstration  of  the  Rule. 


IZT 


Put  the  transverse  axis  ST=a;  the  conjugate  Nw=c  % 
the  height  Ta.  or  Ncr,  of  any  segment =0-,  and  its  cor- 
responding   ordinate    ab—y;    then  by    the  property 

c 

of  the  curve,  y—^^  ax — x2  aud  the   fluxion  of  the 

a  2c      . 

iegment  bTb,   or  bNb= — Xxy/ax — x2.      Now,  ,2x 

a 
^/ax — x*  is  known  to  be  the  fluxion  of  the  correspond- 
ing circular  segment,  BTB  or  bNb.     Therefore,  if  the 
circular  area  be  denoted  be  A,  the  elliptical  area  will 

c 
be    expressed   by — xA.    Hence   it  appears  that  the 

a 
area  of  the  segment  of  the  ellipsis  it  to  that  of  the 
corresponding  circle  as  a  to  c :  and    consequently  the 
whole  ellipsis,  to  the   whole  circle,  in   the  same  pro- 
portion. 


Fig.  1. 


Fig.  2. 


Prom  a  due  consideration  of  the  two  figures,  and 
the  foregoing  demonstrations,  the  following  rule,  for 
finding  the  area  of  an  elliptical  segment^  is  easily  de- 
duced. 


12S  Jle nsuration  of  Superfi c  its. 

Given  the  height  of  an  elliptical  segment,  and  the  two 
diameters,  to  find  the  area. 

RULE. 

1.  Subtract  the  height  of  the  segment,  from  that 
diameter  of  which  it  is  a  part?  multiply  the  remain- 
der by  the  height  of  the  segment,  and  twice  the-square 
root  of  the  product  will  be  the  chord  of  a  circular 
segment,  having  the  same  height. 

2.  Find  the  area  of  this  gegment  by  rule  2,  for  cir- 
cular segments. 

3.  Then,  as  that  diameter,  of  which  the  height  of 
the  segment  is  a  part,  is  to  the  other  diameter;  so  is 
the  circular  area,  to  the  elliptical  area. 

Note.  If  the  chord  of  the  elliptical  segment  also 
be  given ;  then,  instead  of  the  first  part  of  the  above 
rule,  say,  as  that  diameter  to  which  the  chord  of  the 
segmant  is  parallel,  is  to  the  other  diameter,  so  is  the 
chord  of  the  elliptical  segment,  to  the  chord  of  the 
circular  segment,  and  proceed  as  above. 

EXAMPLES. 

1.  In  an  ellipsis,  whose  longer  axis  TS  is  120,  and 
shorter  Nn  40,  what  is  the  area  of  a  segment  thereof, 
6T5,  cut  parallel  to  the  shorter  axis  N«,  the  height  of 
aT  being  24  ? 

TS— T«=Sa=:96  ;  this  multiplied  by  Ta,  24,  gives 
2304;  twice  the  square  root  of  which,  is=96,  the 
chord  BaB,  of  the  circular  segment  BTB,  whose  area 
by  the  second  rule  for  that  purpose,  will  be  found  to 
be  160S.  Then,  as  120 :  40 ;  or,  as  3  to  1 ::  1608 :  536, 
the  area  of  the  elliptic  segment  bTb. 

Note.  If  the  chord  bab  had  been  given  32  ;  BaB 
mighUiave  been  found  thus,  as  40  :  120 ::  32  :  96. 

2.  In  an  ellipsis  whose  longer  axis  TS  is  120,  and 
shorter  Nw40,  what  is  the  area  of  a  segment  thereof, 


Mensuration  of  Superficies.  129 


BNB,  cut  parallel  to  the  longer  axis,  TS,  the  height 
dN  being  4  ? 

Proceeding  as  before,  the  chord,  bob,  of  the  circular 
segment  bNb  {Fig.  2.)  will  be  found=24,  and  its  area 
=  65^-;  therefore,  as  40  :  120;  or  as  1:3  ::  65£:  196,  the 
area  of  the  elliptic  segment  BNB. 
v  3.  Required  the  area  of  an  elliptical  segment  cut 
offby  a  chord  or  double  ordinate,  parallel  to  the  short- 
er diameter,  the  height  of  the  segment  being  10,  and 
the  two  diameters  35  and  25?  Ms.  161.878. 

4.  What  is  the  area  of  an  elliptical  segment  cut 
off  by  a  chord,  parallel  to  the  longer  diameter,  the 
height  of  the  segment  being  5,  and  the  two  diameters 
35  and  25.  Ms.  97.7083. 

5.  What  is  the  area  of  an  elliptical  segment  cut 
off  by  a  chord,  parallel  to  the  shorter  diameter,  the 
height  being  20,  and  the  two  diameters  70  and  50  ? 

Ms.  647.513997. 

Given  the  two  diameters  of  an  Ellipsis  to  find  the  cir- 
cumference. * 

RULE  I. 

Multiply  the  sum  of  the  two  diameters  by  1.5708, 
and  the  product  will  be  the  circumference;  near 
enough  for  most  practical  purposes. 

RULE  11. 

To  half  the  sum  of  the  two  diameters  add  the  square 
root  of  half  the  sum  of  their  squares  5  multiply  this 
last  sum  by  1.5708,  and  the  product  will  be  the  cir- 
cumference, extremely  near. 

EXAMPLES. 

1.  The  two  diameters  of  an  ellipsis  are  24  and  18, 
what  is  the  circumference  ? 


By  rule  1.  n  -j-is 

/21+is        |.^  +  is«\ 
By  rule  2.  f L.^/ Jx  1.3roS=66.30S5, 

the  circumference  of  the  elUf 

-.  W  hat  i»  the  eireumference  of  an  ellipsis  whose 
trau?  »o,  and  conjugate  l 

s-  The  1  ,d  the 

short  v, 

by  rule  1. 
<>6  by  rui 

w  of  an  Elliptical  Ring,  or  the  space 
ided  between  the  Circumferences  of  two  concen- 

RULE. 

Find  the  area  of  each  ellipsis,  and  subtract  the  less 
from  the  greater,  the  remainder  will  be  the  area  of 
the  ring.  Or,  from  the  produet  of  the  two  diameters 
of  the  greater  ellipsis,  subtract  the  produet  of  the 
two  diameters  of  the  less :  the  remainder  multiplied 
bv  .t^34,  will  be  the  area  of  the  ring. 

Xote.  This  rule  will  also  serve  for  a  circular  ring; 
for  when  the  diameters  of  each  ellipsis  become  equal, 
the  square  of  the  diameter  of  the  greater  circle,  di- 
minished by  the  square  of  the  diameter  of  the  less, 
and  the  remainder  multiplied  by  he  area  of 

the  circular  ring:  Or,  multiply  the  sum  of  the  diame- 


*  It  is  here  supposed,  that  the  difference  between  the  conji 
gate  diame  ers  is  equal  to  the  difference  between  the  tr. 
diameters ; 
archt, 

tween  the  semi-transverse,  or  semi-conjugate  diametei 
theler.  i  between  the  arc! 


Mensuration  cf  Superficies. 


131 


ters  by  their  difference,  and  that  product  into  .7854  for 
the  area  of  a  eireular  ring. 


MPLES. 


1.  The  transverse  diame- 
ter AB  of  an  ellipsis  is  60, 
the  conjugate  CD  47;  and 
the  transverse  diameter  EF 
of  another  ellipsis^having 
the  same  centre  Of  is  43, 
and  the  conjugate  GH  32: 
required  the  area  of  the 
space  contained  between 
their  circumferences. 


AB*CD=2820. 
EFXGH=i440. 


Their  difference=l3S0. 

Multiplied  by         .7854 

Gives  the  area  of  the  ring=10S3.S32 

2.  The  ellipsis  in  Grosvenor-square  "measures  840 
links  the  longest  way,  and  612  the  shortest,  within  the 
wall ;  the  wall  is  14  inches  thick,  what  quantity  of 
ground  does  it  inclose,  and  how  much  does  it  stand 
upon  ? 

a       CThe  V'a^  encloses  4  acres  0  roods  60.13  p. 
I  And  stands  on  193.6103  yards. 

^3.  A  gentleman  has  an  elliptical  fountain  in  his  gar- 
den, whose  gr  atest  diameter  is  30,  and  less  24  feet; 
and  has  ordered  a  walk  to  be  paved  round  it  of  3  feet 
jit  inches  in  width,  with  Purbtck  stone,  at  4  shillings 
per  square  yard,  what  will  the  expence  be  ? 


Ms. 


The  area  of  the  walk  is  361.561  feet 
Expence  1.12  :  9s.  :  6d|.  .968. 


LJ2 


Mensuration  of  Superficies. 


4.  The  diameters  of  two  concentric  circles  are  24 
and  18,  what  is  the  area  of  the  space  included  be- 
tween their  circumferences  ?  Jins.  197.9208. 


§  XVI.  To  find  the  JIrea  of  a  Parabola. 

RULE. 

Multiply  the  base  or  greatest  double  ordinate,  by 
the  perpendicular  bright,  and  two  thirds  of  the  pro- 
duct will  be  the  area. 


EXAMPLES. 

1.  If  the  greatest  double  ordinate  GH  be  53.75,  and 
■the  abscissa,  or  height  of  the  parabola  EF,  be  39.25, 
what  is  the  area  ? 

■f  of  53.75X39.25=1406.4583,  the  area  of  the  pa- 
rabola. 


Mensuration  of  Superficies.  13S 

Let  the  altitude  EF  be  denoted  by  x,  and  the  semi- 
ordinate  FH,  by  y,  then  if  a  be  the  latus  rectum  of  the 
parabola,  we  have  from  the  nature  of  the  curve,  a  x 

y*  •     2yy  •    2y*y 

—y*  and  x= — ;  hence  x= ,   and  y  x= -the 

a  a  a 

y3 

fluxion  of  the  areaHEF;  and  its  fluent=fx — =|X 

a 
a?y=fxEFxFH$  therefore   the   area  of  the  whole 
parabola  GEH,  is=f  xGHxEF,  or  two  thirds  of  the 
circumscribing  parallelogram. 

Note.  If  the  ordinate  GH  was  not  at  right  angles 
to  the  axis  EF,  still  the  included  area  would  be  ex- 
pressed by  two  thirds  of  the  base,  multiplied  into  the 
perpendicular  height  of  the  curve  above  said  base  or 
ordinate. 

Demonstration  of  the  rule.  Let  FH,  the  ordi- 
nate, he  divided  into  an  infinite  number  of  equal  parts 
n,  and  suppose  lines  ef,  parallel  and  equal  to  EF,  to 
be  drawn  through  each  of  these  equal  divisions. 
Then,  by  the  nature    of  similar  triangles,  assuming 

±23  n 

EF=1,  ge  will  be  equal  to  o, — j J — ,&e. — to — ; 

n     n      n  n 

and  by  a  property  of  the  parabola. 

(Emerson's  Conies.  B.  III.  Prop.  T.) 


EF=e/    : 

ge    ::  ge 

:    pe 

continually,  viz. 

1             1 

1 

1      : 

—    ::    — 

:     — 

n            n 

rc* 

. 

2              2 

4 

1      : 

n           n 

w2 

3              3 

9 

n* 

1      : 

—     ::     — .     : 

—  &c  — 

—  to  — 

n  .         n 

n% 

/i3 

N 


t34  'Hon  of  Solids. 

Hence  the   area   of   the   space   BKHAE  will   b< 

14      0  n% 

pressed   by  o-| 1 J &c. to — ,  but  the  sum 

.,       n*     n*  n* 

of  this  series  (Emerson's  Arithmetic  of  Difnites,  Prop. 
3)=}w.  Now  the  area  of  the  parallelogram  EKHF 
■=FHx'EF=»Xl  =  n,  and  n — \n~jn.  Hence  the 
I  i'  Ihe  semi-parabola  is  |  of  the  area  of  its  cir- 
eumscribing  parallelogram. 

2.  AVhat  is  the  area  of  a  parabola,  whose,  height  or 
abscissa  is  10 ;  and  the  base,  or  double  ordinate,  16  ? 

Ms.  106f. 

3.  Required  the  area  of  a  parabola  whose  base,  or 
double  ordinate,  is  38,  and  the  height,  or  abscissa,  12  ? 

Ms.  301. 


CHAPTER  II. 


Ilie  MExsvRA<rioK  of  Solids. 

feOLID  bodies  are  such  as  consist  of  length,  breadth, 
and  thickness  \  as  stone,  timber,  globes,  bullets,  &c. 

A  Table  of  Cubic -Measure. 

1728     cubic,  or  solid  inches,  make  1  solid  foot. 

27         -     -     -     -     feet       -     -      1     -       yard. 

166|       -     -     -     -    yards    -     -     1     -      pole. 

64000         ....    poles    -     -     1     -      furlong. 

512         -     -     -     -     furlongs  1     -       mile. 

The  least  solid  measure  is  a  cubic  inch,  and  all 
solids  are  measured  by  cubes,  whose  sides  are  inches, 
feet,  yards,  &e; ;  and  the  solidity  of  a  body  is  said 


Mensuration  of  Solids. 


135 


to  be  so  many  cubie  inches,  feet,  yards,  &c.  as  will 
fill  the  same  space  as  the  solid,  or  as  the  solid  will 
contain. 


§  I.  Of  a  Cube. 

A  ctibe  is  a  solid,  comprehended  under  six  geome,- 
ical  squares,  and  may  be  represented  by  a  die. 


To  find  the  Solidity* 


RULE. 


Multiply  the  side  of  the  cube  into  itself,  and  that 
product  again  by  the  side  5  the  last  product  will  be  the 
solidity, 


EXAMPLES. 


1.  Suppose  ABCDEFG  to  be  a  cubical  piece  of 
stone  or  timber,  each  side  being  17.5  inches,  what  is 
the  solidity  ? 


17.5X17.5x17.5=5359.375  the  solid  content  of  the 
rone  in  inches,  which  divided  by  1728,  gives  3.10149 
ieet. 


136  Mensuration  of  Solids. 

By  Scale  and  Gwnpm 

Extend  the  compasses  from  1  to  17. r>;  that  extent 
turned  over  twice  from  l7.j  will  reach  to  5850,  the 
solid  content  in  inches.  Or,  extend  the  compasses  from 
100  to  17.>,  that  extent,  turned  twice  over  will  reach 
G,  which  multiplied  by  10,  gives  53G0.  Then  ex^ 
tend  the  compasses  from  1728  to  1  ;  that  extent,  turn- 
ed the  same  way,  from  5339,  will  reach  to  3.1  feet. 

Demonstration.  If  the  square 
ABCD  be  conceived  to  be  moved  down 
ihe  plane  ADEF,  always  remaining  pa- 
rallel to  itself,  there  will  be  generated, 
by  such  a  motion,  a  solid,  having  six 
planes,  the  two  opposite  ones  of  which 
will  be  equal  and  parallel  to  each 
other;  whence  it  is  called  a  parallelo- 
pipedon,  or  square  prism.  And  if  the  plane  ADEF 
be  a  square  equal  to  the  generating  plane  ABCD, 
then  will  the  generated  solid  be  a  cube.  From  hence 
such  solids  may  be  conceived  to  be  constituted  of  an 
infinite  series  of  equal  squares,  each  equal  to  the 
square  ABCD;  and  AE  or  DF  will  be  the  number  of 
terms.  Therefore,  if  the  area  of  ABCD  be  multipli- 
ed into  the  number  of  terms  AE,  the  product  is  the 
sum  of  all  the  terms,  and  consequently,  the  solidity  of 
the  parallelopipedon  or  cube.  Or,  if  the  base  ABCD, 
be  divided  into  little  square  areas,  and  the  height  AE 
be  divided  in  a  similar  manner,  you  may  conceive  as 
many  little  cubes  to  be  generated  in  the  whole  solid,  as 
are  equal  to  the  number  of  the  little  areas  of  the  base 
multiplied  by  the  number  of  divisions  which  the  side 
.\E  contains. 

From  this  demonstration  it  is  very  plain,  that,  if 
von  multiply  the  area  of  the  base  of  any  prism  into 
its  length  or  height,  the  product  will  be  its  solidity. — 
Any  solid  figure,  whose  ends  are  parallel,  equal,  and 
similar,  and  its  sides  are  parallelograms,  is  called  a 
Prtsm 


Mensuration  of  Solids.  137 

2.  What  is  the  solidity  of  a  cube,  whose  side  is  19 
feet  4  inches? 

Jlns.  7226  feet  4  inches  5  parts  4  seconds. 

3.  A  cellar  is  to  be  dug,  whose  length,  breadth,  and 
depth,  are  each  10  feet  4  inches  5  how  many  solid  feet 
does  it  contain,  and  what  will  it  cost  digging,  at  1  shil- 
ling per  solid  yard  ? 

0      I  'the  content  is     1103^  feet. 
sins.  £  Expence  l2  .  0 .  10^. 

4.  How  many  three  inch  cubes  may  be  cut  out  of  a 
12  inch  cube?  Jlns.  64. 


§  II.  Of  a  Parailelopipedon. 

A  parallelopipedon  is  a  solid  having  six  rectangu- 
lar sides,  every  opposite  pair  of  which  are  equal  and 
parallel. 

To  find  the  Solidity. 

RULE. 

Multiply  the  breadth  by  the  depth,  and  that  pro- 
duct  by  the  length. 

EXAMPLES. 

1.  Let  ABCDEFG  be  a  parallelopipedon,  or  square 
prism,  representing  a  square  piece  of  timber  or  stone, 
each  side  of  its  square  base  ABCD  being  21  inches : 
and  its  length  AE  15  feet,  required  its  solidity. 

*l „___.,  ™e 


3? 


N   2 


138  Mensuration  of  Solids. 

First,  then,  multiply  21  by  21,  the  product  i*  hi, 
the  area  of  the  base  in  inches;  which  multiply  by  ISO, 
the  length  in  inches,  and  the  product  i^  79380,  the  so- 
lid content  in  inches.  Divide  the  lust  product  by 
and  ihe  quotient  is  13.9,  that  is,  45  solid  feet 
and  9  tenths  of  a  foot. 

Or  thus,  by  multiplying  feet  and  inches. 

Multiply  l  foot.  9  inches  by  l  foot  9  inches,  and  the 
product  is  S  feet  0  inches  9  parts  ;  this  multiplied  a- 
gain  by  15  feet,  gives  45  feet  11  inches  3  parts. 

By  Scale  and  Compasses. 

Extend  the  compasses  from  12  to  21  (the  side  of  the 
square)  and  that  extent  will  reach  to  near  46  feet,  be- 
ing twice  turned  over  from  15  feet;  so  the  solid  con- 
tent  i*  almost  46  feet. 

Note.  When  the  breadth  and  depth  of  the  solid 
are  unequal,  a  mean  proportional  between  them  must 
be  found,  by  dividing  the  space  between  them  into  two 
equal  parts,  for  the  side  of  a  mean  square  in  inches  ; 
then  proceed  as  above.  For,  as  12  is  to  the  side  of  a 
mean  square  in  inches,  so  is  the  length  in  feet  to  a 
fourth  number;  and  so  is  this  fourth  number  to  the 
content  in  foot  measure,  of  any  parallelopipedon. 

2.  If  a  piece  of  timber  be  25  inches  broad,  9  inches 
deep,  and  25  feet  long,  how  many  solid  feet  are  con- 
tained therein  ?  Jlns.  39  feet  0  inches  9  parts. 

3.  If  a  piece  of  timber  be  15  inches  square  at  the 
end,  and  18  feet  long,  how  many  solid  feet  are  con- 
tained in  it  ;  and  suppose  I  want  to  cut  off  a  solid  foot 
from  one  end  of  it,  at  what  distance  from  the  end 
mustl  cut  it? 

q        C  The  solidity  is  28.125  feet.. 

I  And  7.68  inches  in  length  will  make  1  foot. 

4.  If  a  piece  of  squared  timber  be  2  feet  9  inches 
long,  1  foot  7  inches  broad,  and  16  feet  9  inches  long, 


Mensuration  of  Solids. 


139 


how  many  feet  of  timber  are  in  that  piece ;  and  how 
much  in  length  will  make  a  solid  foot  ? 

The  solid  content  is  72.93  feet,  or  72  feet  11 
Ans.  <{      inches  2  parts  3  seconds;  and  2.75%  inches 


'•{! 


in  length  make  1  solid  foot. 


§  III.  Of  a  Prism. 

A  prism  is  a  solid  contained  under  several  planes, 
and  having  its  bases  similar,  equal,  and  parallel. 

To  find  the  Solidity. 

RULE. 

Multiply  the  area  of  its  base  or  end,  by  the  height 
or  length,  and  the  product  will  be  the  solidity. 


EXAMPLES. 

1.  Let  ABCDEF  be  a  triangu- 
l  :ism,  each  side  of  the  base  be 


ing  10.6  inehes,  the  pcrpeadicular 
Ca  13.51  inches,  and  the  length  of 
the  solid  19.5  feet.  How  many 
solid  feet  are  contained  therein  ? 

Multiply  the  perpendicular  of 
the  triangle  13.51  by  half  the  side 
7.8,  and  the  product  is  105.378, 
the  area  of  the  base ;  which  mul- 
tiply by  the  length  19.5,  and  the 
product  is  2054.871  ;  which  divide 
by  144,  and  the  quotient  is  14,27 
feet  fere,  the  solid  content. 


Mensuration  of  Solids. 


By  Scale  and  Compasses. 

First,  find  a  mean  proportional  between  the  perpen- 
dicular and  half  side,  by  dividing  the  space  upon  the 
line,  between  13.51  and  7. 8  into  two  equal  parts;  so  shall 
you  find  the  middle  point  between  them  to  be  at  10.26, 
which  is  the  mean  proportional  sought :  by  this  means 
the  triangular  solid  is  brought  to  a  square  one,  each 
side  being  10.26  inches.  Then  extend  the  compasses 
from  ]2  to  0.26  ;  that  extent,  turned  twice  downwards 
from  19.5  feet,  the  length,  will  at  last  fall  upon  14.27, 
which  is  14  feet  and  a  little  above  a  quarter. 


2.  Let  ABCDEFGHTK 
represent  a  prism,  whose  base 
is  a  hexagon,  each  side  be- 
ing 16  inches,  the  perpendic- 
ular from  the  centre  of  the 
base  to  the  middle  of  one  of 
the  sides  (ab)  13.84  inches, 
and  the  length  of  the  prism 
15  feet  :  the  solid  content  is 
required. 

Multiply  half  the  sum  of 
the  sides  48  by  13.84,  and 
the  product  is  864.32,  the 
area  of  the  hexagonal  base 
(by  §  VIII.  p.  96.)  which 
multiplied  by  15  feet,  the 
length,  the  product  is  9964.8; 
which  divided  by  14*,  the 
quotient  will  be  6  9.2  feet, 
the  solid  content  required. 


Mensuration  of  Solids.  141 


By  Scale  and  Compasses. 

First,  find  a  mean  proportional  between  the  perpen- 
dicular, and  half  the  sura  of  the  sides ;  that  is  divide 
the  space  between  13.84  and  48,  and  the  middle  point 
will  be  25.77,  the  side  of  a  square  equal  in  area  to 
the  base  of  the  solid.  Then  extend  the  compasses 
from  12  to  25.77 ;  that  extent  will  reach  (being  twice 
turned  over)  from  15  feet,  the  length,  to  69.2  feet,  the 
content. 

To  find  the  Superficial  Content  of  any  Prism. 

RULE, 

Multiply  the  circumference  of  the  base  by  the  length 
of  the  prism ;  the  product  will  be  the  upright  surface*  5 
to  which  add  the  area  of  the  bases  3  the  sum  will  be 
the  whole  superficial  content. 

EXAMPLES. 

In  the  hexagonal  prism  last  mentioned,  the  sum  of 
the  sides,  or  circumference  of  the  end,  being  96,  and 
the  length  15  feet,  that  is,  180  inches :  which  multi- 
plied by  96,  the  product  is  17280  square  inches ;  to 
which  add  twice  664.32,  the  areas  of  the  two  bases, 
the  sum  is  18608.64,  the  area  of  the  whole,  which  is 
129.226  feet. 

1.  Each  side  of  the  base  of  a  triangular  prism  is  ft 
inches,  and  the  length  12  feet  5  inches  5  what  is  the 
solidity  and  surface  ? 


a       J  The  solidity  10.79  feet 
'  \  Surface  54.50S9  feet 


142  Mensuration  of  Solids. 

\  hexagonal  prism  measures  28  inches  across  (ho 
'(ntre  of  thf  end,  from  corner  to  corner,  and  ■ 
a  in  length;  required  the  solidity  and  surface. 

a        J  The  solidity  is  39.48835  feet. 
£  Surface  85.2893  feet. 

A  decagonal  prism,  or  pillar,  measures  50  inches 
rcumferenee,  and  is  30  feet  high;  required  the 
solidity  and  surface. 

q       ^  The  solidity  40.074  feet. 
"*•  £  Surface  137.6718  feet. 

4.  The  gallery  of  a  church  is  supported  by  10  oeta* 
gonal  prisms  of  wood,  which  measure  each  48  inches 
in  circumference,  and  are  each  12  feet  highj  what  will 
he  the  expence  of  painting  them  at  10  pence  per 
square  yard  ?  dns.  1.2  :  7s  :  9i. 

5.  A  trapezoidal  prism  of  earth,  or  part  of  a  canal, 
is  to  he  dug,  whose  perpendicular  depth  is  10  yards, 
the  width  at  the  top  20  yards,  at  the  bottom  16,  and 
the  length  50  yards,  the  two  ends  being  cut  perpendi- 
cularly down ;  how  many  solid  yards  of  earth  are  con- 
tained in  this  part  ?  %ins.  9000  yards. 


£  §IV.  OfaPrkAMiD. 

A  pyramid  is  a  solid  figure,  the  base  of  which  is  a 
polygon,  and  its  sides  plain  triangles,  their  several 
vertical  angles  meeting  together  in  one  point. 


Mensuration  of  Solids. 


14ci 


To  find  the  Solidity. 

RULE. 


Multiply  the  area  of  the  base  by  a  third  part  of  the 
altitude,  or  length  5  and  the  product  is  the  solid  con- 
tent of  the  pyramid. 


Example  1.  Let  ABD  be  a 
square  pyramid,  having  each 
side  of  its  base  18  5  inches, 
and  the  perpendicular  height 
CD  15  feet,  what  is  the  soli- 
dity? 

Multiply  13.5  by  18.5,  and 
the  product  is  342,25,  the  area 
of  the  base,  in  inches,  which 
multiply  by  5,  a  third  part  of 
the  height,  and  the  product  is 
1711.25 ;  this  divided  by  144 
gives  11.83  feet,  the  solid  con- 
tent. 


To  find  the  Superficial  Content 


rule. 


Multiply  the  slant  height  by  half  the  circumference 
of  tike  base,  and  the  product  will  be  the  upright  sur- 
face.— To  which  the  area  of  the  base  may  be  added, 
for  the  whole  surface 


.Mensuration  of  Solids. 

Note.  This  rule  will  serve  for  the  surface  of  all 
pyramids.  Perhaps  it  may  he  proper  here  to  acquaint 
the  learner,  that  the  slant  height  of  any  pyramid  is 
not  the  height  from  one  of  the  corners  of  the  hose  to 
the  vertex  or  top,  hut  from  the  middle  of  one  side  of 
the  base.  And  the  perpendicular  height  of  a  pyramid, 
is  a  line  drawn  from  the  vertex,  to  the  middle  or  centre 
of  the  base;  henee  it  will  he  necessary  to  find  the  dis- 
tance between  the  centre  of  the  hase  of  a  pyramid, 
and  the  middle  of  one  of  the  sides. — This  distance 
may  always  be  found  by  multiplying  the  tabular  per- 
pendicular in  section  V.  p.  84,  by  one  of  the  sides 
of  the  base;  then,  if  to  the  square  of  this  number, 
you  add  the  square  of  the  perpendicular  height  of  the 
pyramid,  the  square  root  of  the  sum  will  give  the 
slant  height. 

EXAMPLES. 

1.  Required  the  surface  of  the  foregoing  pyramid. 

To  the  square  of  the  perpendicular  height  De  15 
feet  or  180  inches,  add  the  square  of  de  9.25,  the  dis- 
tauce  from  the  centre  e,  of  the  base,  to  the  middle  d  of 
one  of  the  sides  ;  which,  in  a  square  base,  is  always 
equal  to  half  the  side.  The  square  root  of  the  sum, 
viz.  32485.5625  is  180.24  nearly,  the  slant  height  Drf. 
IV ow  half  the  circumference  of  the  base  is  37,  which 
multiplied  by  180.24,  gives  666S.88  inches  for  the  up- 
right surface ;  to  which  add  342.25  the  area  of  the 
base,  the  sum  is  7011.13  inches,  the  whole  surface 
equal  to  48.69  feet  nearly. 

Demonstration  of  the  rule.  Every  pyramid  is 
a  third  part  of  a  prism,  that  has  the  same  base  and 
height  (by  Euclid,  XII.  7.) 

That  is,  the  solid  content  of  the  pyramid  ABD  (in 
the  last  figure)  is  one  third  part  of  its  circumscribing 
prism  ABEF.    Now  the  solidity  of  a  prism  is  found 


Mensuration  of  Solids. 


145 


by  multiplying  the  area  of  the  base  into  the  height; 
therefore  the  solidity  of  a  pyramid  will  be  found  by 
multiplying  the  area  of  the  base  by  the  height,  and 
taking  one  third  of  the  product.  You  may  very  easily 
prove  a  triangular  pyramid  to  be  a  third  part  of  a 
prism  of  equal  base  and  altitude,  mechanically,  by 
making  a  triangular  prism  of  cork,  and  then  cutting 
that  prism  into  three  Pyramids,  in  a  diagonal  direc- 
tion. 

2.  Let  ABCD  be 
lar  pyramid,  each  side  of  the 
base  being  21.-5  inches,  and  its 
perpendicular  height  16  feet  5 
required  the  solidity  and  sur- 
face. 

First,  the  area  of  the  base, 
by  sect.  V.  is  200.19896  inches, 
which  multiplied  by  64?,  the 
third  part  of  the  height  in 
inches,  gives  12812.73344  inch- 
es, the  solidity,  :qual  to  7.41477 
feet.  Again,  the  distance  from 
the  centre  of  the  triangle  ABC 
to  the  middle  of  one  of  the  sides, 
as  d  in  the  side  AC,  is  6.20652 ; 
to  the  square  of  this  number, 
which  is  38.5208905104,  add 
the  square  of  the  height  36864 
inches,  the  square  root  of 
the  sum  36902.5208905104,  is 
192.10028  inches,  the  slant  AJ 
height  dD.  Hence  the  upright 

surface  is  6195.23403  inches,  and  the  whole  surface 
6395.43299  inches,  or  44.41272  feet. 

O 


H6 


JMenmration  of  Solids. 


8.  Let    1BCDEFGH  be  s 
pyramid,  whose  base  is  a  hep- 

■ach  side  of  it  being  1/5 

inches,  ami  the   perpendicular 

heigh*!    of    the    pyramid.    Ill, 

ee< ;  required  the  solidity 

and  superficies. 

First,  1.3  multiplied  by 
1.0382017,  the  tabular  perpen- 
dicular for  a  heptagon  (sect. 
VIII.)  gives  15.573<)255  inches, 
the  distance  from  the  centre  1 
to  the  middle  of  the  side  AB; 
this  multiplied  by  52.5,  half 
the  sum  of  the  sides  of  the 
base,  gives  817.63108873  inch- 
es, the  area  of  the  base.  This 
last  number  multiplied  by  4.3, 
one  third  of  the  height,  and 
divided  by  144,  will  give 
feet,  the  solidity. — Again,  if  to 
the  square  of  15.5739255,'  vou 
add  the  square  of  the  height 
in  inches,  the  square  root  of  the 
sum  26486.5471." 31 8  will  be  162.74688,  the  slant  height 
of  the  pyramid  5  this  multiplied  by  52.5  gives  8544.21 ti 
inches,  the  upright  surface,  to  which  add  the  area  oi 
the  base,  and  the  whole  surface  will  be  9361.8422887* 


inches,  or  65.0128  feet. 


By  Scale  and  Compasses, 

First,  find  a  mean  proportional  between  15.57  an< 

52.5,  by  dividing  the  space  between  them  into  twc 
equal  parts,  and  you  will  find  the  middle  point  to  be 

28.6,  the  side  of  a  square  equal  in  area  to  the  base  oi 
the  pyramid;  then  extend  the  compasses  from  12  to 
28.6,  that  extent  will  reach  from  4.5  (twice  turned 
over)  to  25.55  feet,  the  solid  content.  Or,  extend  the 
compasses  from  1  to  the  area  of  the  base,  that  extent 


Mensuration  of  Solids.  147 

will  reach  from  one-third  of  the  height  to  the  solidity  ; 
the  dimensions  being  all  of  the  same  name. 

4.  There  is  a  triangular  pyramid,  the  sides  of  its 
base  are  13,  14,  and  15  feet,  and  its  altitude  63  feet, 
what  is  the  solidity  ?  Ms.  1764  feet.  . 

5.  Eacli  side  of  the  base  of  a  hexagonal  pyramid  is 
10,  and  the  perpendicular  height  43,  what  is  the  soli- 
dity and  superficies  ? 

q        C  Solidity  3897.1143  feet. 
.    '  \  Superficies  1634.580328  feet. 

6.  Required  the  solidity  and  surface  of  a  square  py- 
ramid, each  side  of  the  base  being  3  feet,  and  the  per- 
pendicular height  24  feet. 

-        C  Solidity  72  feet. 
*aw{"  I  Surface  153.2808  feet. 

7.  A  square  pyramidal  stone,  whose  slant  height  is 
21  feet,  and  each  side  of  its  square  base  30  inches,  is 
to  be  sold  at  7s.  per  solid  foot ;  and  the  polishing  of 
the  upright  surface  will  cost  8d.  per  foot ;  what  will 
be  the  expence  of  the  stone  when  finished  ? 

Ms.  Us  :  15s. :  8^d. 

8.  The  spire  of  a  church  is  an  octagonal  pyramid 
(built  of  stone)  each  side  of  the  base  being  5  feet  10 
inehes,  and  its  perpendicular  height  45  feet ;  also  each 
side  of  the  cavity,  or  hollow  part,  at  the  base  is  4  feet 
11  inches,  and  its  perpendicular  height  41  feet ;  I  de- 
sire to  know  how  many  solid  yards  of  stone  the  spire 
contains  ? 

("Solidity  of  the  whole  2464.509614  feet. 
Ms.  1  The  cavity  1595.180393  feet. 

|_And  the  stone-work  32.19738  yards. 


§  V.  Of  a  Ctlisder. 

A  cylinder  is  a  solid,  having  its  bases  circular,  equal, 
and  parallel,  in  form  of  a  rolling  stone. 


Ml 


Mensuration  of  Solid*. 
To  find  the  Solidity. 


RULE. 


Multiply  the  area  of  the  base  by  the  length,  and  the 
product  is  the  solid  content. 


EXAMPLES. 


l.  Let  ABC  be  a  cylinder, 
whose  diameter  AB  is  21.5  inch- 
es, and  the  length  CD  is  16  feet, 
the  solid  content  is  required. 

First,  sqnare  the  diameter 
.21.3,  and  it  makes  462.2") ; 
which  multiply  by  .7854,  and 
the  product  is  363.03113:  then 
multiply,  this  by  16,  and  the 
product  is  5808.8184.  Divide 
this  last  product  by  144,  and 
the  quotient  is  40.339  feet,  the 
solid  content 


By  Scale  and  Compasses. 


Extend  the  compasses  from  13.54  to  21.5,  the  dia- 
meter, that  extent  (turned  twiee  over  from  16,  the 
length)  will  at  last  fall  upon  40.34,  the  solid  content. 

Note.  13.54  is  the  diameter  of  a  circle  whose  area 
is  144  inches     and  >  to  the  diameter  of  any 

eylinders  base  in  inches,  so  is  the  length  of  that  cy- 
linder, in  feet,  to  a  fourth  number;  and  so  is  this 
fourth  number,  to  the  solid  content  of  the  cylinder  in 
feet. 

When  the  circumference  is  given  in  inches,  and  the 
length  in  feet,  extend  from  42.54  to  the  circumference 
of  the  cylinder,  that  extent  will  reaehfrom  the  length, 


Mensuration  of  Solids.  149 

turned  twice  over,  to  the  content  in  feet.  For.  as  42.54 
is  to  the  circumference  of  any  cylinder,  in  inches,  so 
is  the  length  of  that  cylinder  in  feet  to  a  fourth  num- 
ber; and  so  is  this  fourth  number  to  the  solidity  of 
the  cylinder  in  feet. 

42.54  is  the  circumference  of  a  circle  whose  area 
is   144. 

To  find  the  Superficial  Content. 

First,  (by  chap.  I.  sect.  IX.  prob.  1,)  find  the  circum- 
ference of  the  base  67.54,  which  multiplied  by  16,  the 
product  is  1080.64;  and  this  divided  by  12,  the  quo- 
tient is  90.05  feet,  the  upright  surface :  to  which  add 
5.04  feet,  the  sum  of  the  areas  of  the  two  bases,  aud 
the  sum  is  95.09  feet,  the  whole  superficial  content. 

2.  If  a  piece  of  timber  be  96  inches  in  circumfe- 
rence, and  18  feet  long,  how  many  feet  of  timber  are 
contained  in  it,  supposing  it  to  be  perfectly  cylindri- 
cal ?  Jns.  91.676  feet. 

3.  If  a  piece  of  timber,  perfectly  cylindrical,  be  86 
inches  in  circumference,  and  20  feet  long:  how  many 
solid  feet  are  contained  therein?     Jlnj±  81.74634  feet. 

4.  The  diameter  of  the  base  of  a  cylinder  is  20.75 
inches,  and  its  length  4  feet  7  inches ;  what  is  its  so- 
lidity and  surface  r  C  Solidity  10.7633  feet. 

I  Surface  29.595  feet. 

5.  I  have  a  rolling  stone,  44  inches  in  circumfe- 
rence, and  am  to  cut  off  3  cubic  feet  from  one  end; 
whereabouts  must  the  section  be  made  ? 

.     At  33.64  inches, 

6.  A  person  bespoke  an  iron  roller  for  a  garden,  the 
outside-  diameter  to  be  20  inches,  the  thickness  of  the 
metal  i\  inch,  and  length  of  the  roller  50  inches; 
now  supposing  every  cubic  inch  to  weigh  41-  ounces, 
what  will  the  whole  come  to  at3id.  per  pound  ? 

f  Solidity  4358.97  inches. 
Ans.  <  Weight  1157.8514  pounds. 
(Co8t7.  15:13s.  :  7d. 
o  2 


150 


Mensuration  of  Solids, 


§  VI.     Of  a  Cone, 

A  cone  is  a  solid,  having  a  circular  base,  and  grow- 
ing proportionally  smaller,  till  it  ends  in  a  point, 
called  the  vertex,  and  may  be  nearly  represented  by  a 
sugar  loaf. 

2b  find  the  Solidity. 

RULE. 

Multiply  the  area  of  the  base  by  a  third  part  of  the 
perpendicular  height,  and  the  product  is  the  solid  con- 
tent. 


EXAMPLES. 


1.  Let  ABC  be  a  cone,  the  di- 
ameter of  whose  base  AB  is  26.5 
inches,  and  the  height  DC  16.5 
feet :  required  the  solidity. 

First,  square  the  diameter  26.5, 
and  it  is  702.25 ;  this  multiplied 
by  .7854  gives  551.54715  inches, 
the  area  of  the  base ;  which  mul- 
tiplied by  5.5,  one  third  of  the 
height,  gives  3033. 509325;  this 
divided  by  144  gives  21.066  feet, 
the  solid  content. 


By  Scale  and  Compasses. 

Extend  the  compasses  from  13.54  to  26.5,  the  diame- 
ter, that  e  ent  turned  twice  over  from  5.5  (a  third 
part  of  the  height,)  it  will  at  last  fall  upon  21.06  feet, 
the  content. — See  the  remark  on  the  number  13.54,  in 
section  V. 


Mensuration  of  Solids,  151 

To  find  th$  Superficial  Content 

The  rule  given  in  sect.  IV.  for  finding  the  superficies 
of  a  pyramid,  will  serve  to  find  the  superficies  of  a  cone. 

First,  find  the  slant  height  thus  :  To  the  square  of 
the  radius,  13. 5,  which  is  ±75.5625,  add  the  square  of 
the  height  in  inches,  39204,  the  square  root  of  the 
sum,  39379.5625,  is  198.4428,  the  slant  height.  The 
radius  of  the  base,  13.25,  multiplied  by  3.1416  gives  half 
the  circumference  of  the  base  41.6262,  and  this  pro- 
duct multiplied  by  the  slant  height,  gives  8260.4196S136 
for  the  upright,  or  convex  surface ;  to  which  add  the 
area  of  the  base  551.54715,and  the  sum  is  8811.96683136 
inches,  the  whole  surface,  equal  to  61.194  feet. 

Demomstration.  Every  cone  is  the  third  part  of 
a  cylinder  of  equal  base  and  altitude  (by  Euclid,  XII. 
10.)  The  truth  of  this  may  easily  be  conceived,  by 
only  considering,  that  a  cone  is  but  a  round  pyramid ; 
and  therefore  it  must  needs  have  the  same  ratio  to  its 
circumscribing  cylinder,  as  a  multangular  pyramid 
hath  to  its  circumscribing  prism,  viz.  as  1  to  3.  For, 
the  base  of  the  multangular  pyramid  may  consist  of 
such  a  number  of  sides,  that  the  difference  between  its 
circumference,  and  that  of  the  circle,  will  be  less  than 
any  assignable  magnitude. 

The  curve  superficies  of  every  right  cone,  is  equal 
to  half  the  rectangle  of  the  circumference  of  its  base 
into  the  length  of  its  side. 

For  the  curve 
surface  of  every 
right  cone  is  e- 
qual  to  the  sec- 
tor of  a  circle, 
whose  arch  BC  is 
equal  to  the  peri- 
phery of  the  base  \  / 
of  the  cone,  and 
radius  AB  equal 
to  the  slant  side  of 
the  cone:   which 


/ 
/ 


Mensuration  of  Solids. 

will  appear  very  evident,  if  you  cut  a  piece  of  paper 
in  the  form  of  a  sector  of  a  circle,  as  ABC,  and  bend 
the  aides  AB  and  AC  together, till  they  meet,  and  you 
will  find  it  to  form  a  right  cone. 

I  have  omitted  the  demonstrations  of  tin1  superfi- 
cies of  all  the  foregoing  solids,  because  I  thought  it 
needless,  they  being  all  composed  of  squares,  paral- 
lelograms, triangles.  &e.  which  figures  arc  all  demon- 
strated before.  And  if  the  area  of  all  such  figures  as 
compose  the  surface  of  the  solid,  be  found  separately, 
and  added  together,  the  sum  Mill  be  the  superficial 
content  of  the  solid. 

2.  The  diameter  of  the  hase  of  a  cone  is  10,  and 
its  perpendicular  height  6S.1 ;  required   the    solidity 


and  superficies. 


a        5  Solidity  1782.858. 

I  Superficies  1151.134076. 

3.  What  is  the  solidity  of  an  elliptical  cone,  the 
greater  diameter  of  its  base  being  15.2,  the  less  10, 
and  the  perpendicular  height  22  ?  dns.  875.4592. 

4.  What  is  the  solidity  and  superficies  of  a  cone, 
whose  perpendicular  height  is  10.5  feet,  and  the  cir- 
cumference of  its  base  9  feet  ? 

a       S  Solidity  22.56093  feet. 
X  Superficies  54.1336  feet. 

5.  What  is  the  solidity  and  superficies  of  a  cone, 
whose  slant  height  is  32  feet,  and  the  circumference  of 
its  hase  24  feet  ? 

a       J  Solidity  4S5.4433316  feet. 
'  i  Surface  429.83804  feet. 

6.  What  will  the  painting  of  a  conical  church  spire 
come  to  at  8d.  per  yard;  supposing  the  circumference 
of  the  base  to  be  61  feet,  and  the  perpendicular  height 
118  feet  rw 

-  •    $  Superficies  3789.76  feet. 
%ms-  ^Expence  J.14  :  Os.  :  bU, 


Mensuration  of  Solids*  153 


§  VII.  Of  the  Frustum  of  a  Ptramid. 

A  frustum  of  a  pyramid  is  the  remaining  part,  when 
the  top  is  cut  off  by  a  plane  parallel  to  the  base. 

To  find  the  Solidity. 


GENERAL     RULE. 

Multiply  the  areas  of  the  two  bases  together,  and  to 
the  square  root  of  the  product  add  the  two  areas; 
that  sum,  multiplied  by  one  third  of  the  height,  gives 
the  solidity  of  any  frustum. 

rule  11. 

If  the  Bases  be  Squares. 

To  the  rectangle  (or  product)  of  the  sides  of  the  two 
bases  add  the  sum  of  their  squares :  that  sum,  being 
multiplied  into  one-third  part  the  frustum's  height,  will 
give  its  solidity. 

RULE    III. 

If  the  Bases  be  Circles. 

To  the  product  of  the  diameters  of  the  two  bases 
add  the  sum  of  their  squares  5  this  sum,  multiplied  by 
the  height,  and  then  by  .2618,  or  one-third  of  .7854^ 
the  last  product  will  be  the  solidity. 

RULE  IV. 

If  the  Bases  be  regxdur  Polygons. 

Add  the  square  of  a  side  of  each  end  of  the  frus- 
tum, and  the  product  of  those  sides  into  one  sum; 
multiply  this  sum  by  one-third  of  the  tabular  area  be- 


154 


Mensuration  of  Solids. 


longing  to  the  polygon  (sect.  VIIT.  p.  99.)  and  this 
product  by  the  height,  for  the  solidity. 


EXAMPLES. 

1.  Let  ABCD  he  the  frus- 
tum of  a  square  pyramid, 
each  side  of  the  greater  base 
18  inches,  each  side  of  the 
Jo**  12  inches,  and  tbe  height 
18  feet;  the  solidity  is  re- 
quired. 


BY  THE  GENERAL  RULE. 

The  square  of  18  is  324, 
and  the  square  of  42  is  141; 
also  the  product  of  these  two 
squares  is  46656,  the  square 
root  of  which  is  216,  a  mean 
area. 

Area   of  the    greater 
base      -     -     -     -       324 

Area  of  the  less  base      144 

Mean  area    -    -     -     216 


,'XI 


» 


Sum  -  6S4  which  multiplied  by  6,  one 
third  of  4he  height,  gives  4104;  this  divided  by  144 
gives  28.5  feet  the  solidity. 


BY  RULE  2. 


Product  of  the  sides  of  the  bases  18  and  12  is 
Square  of  the  greater  side  18  is  -  -  -  - 
Square  of  the  less  side  12  is 


Then  proceed  as  above. 


Sum 


216 
324 
144 

684 


Mensuration  of  Solids.  4  >;> 

To  find  the  Superficial  Content. 


Multiply  half  the  sum  of  the  perimeters  of  the  two 
bases  by  the  slaut  height,  and  to  that  product  add  the 
areas  of  the  two  bases  for  the  whole  superficies. 

Note.  The  slant  height  of  any  frustum,  whose  ends 
are  regular  polygons,  is  a  line  drawn  from  the  middle 
of  one  side  of  the  less  end,  to  the  middle  of  its  paral- 
lel side  at  the  greater  end.  And,  the  perpendicular 
height,  is  a  line  drawn  from  the  centre  of  the  less  end, 
to  the  centre  of  the  greater;  or  it  is  a  perpendicular 
let  fall  from  the  middle  of  one  of  the  sides  of  the  less 
end,  upon  the  surface  of  the  greater  end.  Hence  the 
slant  height,  and  perpendicular  height,  will  be  two 
sides  of  a  right  angled  triangle ;  the  base  of  which 
will  be  equal  to  the  difference  between  the  radii  of  the 
inscribed  cireles  of  the  two  ends  of  the  frustum.  And 
this  base  may  always  be  found,  by  multiplying  the  dif- 
ference between  a  side  at  each  end  of  the  frustum,  by 
the  tabular  perpendicular  in  section  VIII.  p.  100.  The 
perpendicular  height  of  the  pyramid  of  which  any 
frustum  is  a  part,  may  readily  be  found,  by  saying,  as 
the  base  found  above,  is. to  the  perpendicular  height 
of  the  frustum  5  so  is  the  radius  of  the  inscribed  cir- 
cle of  the  frustum's  base,  to  the  perpendicular  height 
of  the  pyramid.  The  radius  of  the  inscribed  circle 
is  found  by  multiplying  a  side  of  the  base,  by  the  tab- 
ular perpendicular,  section  VIII. 

Example.  Required  the  superficies  of  the  foregoing 
frustum. 

The  perimeter  of  the  greater  base  72  inches. 
The  perimeter  of  the  less  base  is    48  inches. 

Sum     -     -     120 

Half  sum    -    -      00 


i56 


Mensuration  of  Solids. 


From  19,  a  side  of  the  greater  end,  take  12,  a  side 
of  the  less;  the  difference  6,  multiplied  by  the  tabular 
perpendicular  .5,  gives  3.  The  perpendicular  height 
of  the  frustum  in  inches  is  216  :  to  the  square  of  216, 
which  is  46656,  add  the  square  of  3,  which  is  9,  the 
square  root  of  the  sum  46656  is  216.02083  inches,  the 
giant  height,  which  multiplied  by  60  gives  12^61.2498 
inches.  To  this  product  add  the  areas  of  the  two  ends 
324,  and  144,  and  the  whole  surface  will  be  13429.2498 
inches,  or  93.2586  feet. 

Example  2.  Let  ABCD  be 
the  frustum  of  a  triangular  py- 
ramid, each  side  of  the  greater 
base  25  inches,  each  side  of  the 
less  base  9  inches,  and  the  length 
15  feet ;  the  solid  content  of  it 
is  required. 


BY  RULE  4. 


The  square  of  25  is 
The  square  of  9  is 
Product  of  25  and  9  is 


625 

81 
225 


Sum  931 

The  tabular  area  is  .433013, 
one-third  of  which  is  .1443377,  CI 
this  multiplied  by  931,  and  then 
by  15,  produces  2015.6759805,  which  divided  by  1443 
the  quotient  is  13.9977  feet,  the  solidity. 


OR  THUS,  BY  THE  GENERAL  RULE. 


The  square  of  25,  multiplied  by  the  tabular  area, 
(section  VIII.)  gives  the  area  of  the  greater  base 
270.63312.^:  in  a  similar  manner  the  area  of  the  less 
base  wiirbe  found  to  be  35.07405 ;  the  square  root  of 
the  product  of  these  two  areas  9492.200569805625  is 
97.427U25. 


Mensuration  of  Solids.  157 

Area  of  the  greater  base     -       -         -  270.633125 

Area  of  the  less  base         -  35.074053 

Mean  proportional  between  them      -  27.427925 


Sum         -         -         403.135103 

which  multiplied  by  5,  one  third  of  the  height,  and  di- 
vided by  144,  gives  13.9977  feet,  the  solidity  as  be- 
fore. 


To  find  the  Superficial  Content. 

The  perimeter  of  the  greater  base  is      75   inches. 
The  perimeter  of  the  less  base  is     -      27   inches. 

Sum      -        102 

Half  sum        -         51 


From  25,  a  side  of  the  greater  end,  take  9,  a  side  of 
the  less,  the  difference  16,  multiplied  by  the  tabular 
perpendicular  .2886751,  gives  4.6 1 880  i  6. — The  per- 
pendicular height  of  the  frustum  in  inches,  is  180; 
to  the  square  of  180,  which  is  32400,  add  the  square  of 
4.6188016,  which  is  21.33332822,  is  180.0592  inches, 
the  slant  h.  ight  ;  which  multiplied  by  51  gives 
9183.0192  inches.  To  this  product  add  the  areas  of 
the  two  ends  270.63312  -,  and  35.074053,  and  the  whole 
surface  will  be  9488.726378  inches,  or  65.893y33  feet. 


Mem 


\\\(  \)  to  be  id" 
frustum  of  a  pyramid, 
having  an  octagonal 
base.  i-a<  \\  suit*  of  tt 
being  9  inches,  each 
side  of  tlie  less  1) 
5  inches,  ami  the 
length,  or  height,  10.8 
feet ;  the  solidity  is 
required. 

by   rule  4. 

The  square  of  9  is  81 
The  square  of  5  is  25 
Product  of  5  and  A 

9  is       -     -  45 


>um     - 


151 


Tabular  area  is  4.828427,  one  third  of  which  is 
1.6094756  ;  this  multiplied  by  151,  and  then  by  10.5, 
produces  2551.8235638,  which  divided  by  1.44,  the  quo- 
tient, is  17.721  feet,  the  solidity. 


To  find  the  Superficial  Content. 


72  inches. 
40 

112 

56 


The  perimeter  of  the  greater  base  is 
The  perimeter  of  the  less  base  is 

Sum     - 

Half  sum     - 

From  9,  a  side  of  the  greater  base  or  end,  take  5,  a 
side  of  the  less,  the  remainder  4  multiplied  by  the  tabu- 
lar perpendicular  1.2071068,  gives  4.8284272.  The 
perpendicular  height  of  the  frustum  in  inches  is  126  ; 
to  the  square  of  126,  which  is  15876,  add  the  square  of 


/  i  \ 


Mensuration  of  Solids.  159 

4.8284272,  which  is  23.3137092257,  the  square  root  of 
the  sum  15899.3137092237,  is  126.09248,  the  slant 
height ;  which  multiplied  by  56  gives  7061.17888  inch- 
es. To  this  product  add  the  areas  of  the  two  ends, 
391.102587,  and  120.710675  (found  by  rule  2,  section 
VIII.)  and  the  whole  surface  will  be  7572.992142  in- 
ches, or  52.590223  feet. 

Demonstration.  From  the  rules  delivered  in  the 
IVth  and  Vlth  sections,  the  preceding  rules  may  easi- 
ly be  demonstrated. 

Suppose  a  pyramid  ABY,  to  he 
cut  by  a  plane  at  ab9  parallel  to  M 

its  base  AB,  and  it  were  required  /  j  \ 

to  find  the  solidity  of  the  frustum, 
or  part  AaoB. 

i       i 
Let  D=AB,  a  line,  or  diameter       , 
of  the  greater  base.  h 

d=.ab,  a  similar  line  of  the        l\ 

less  base. 
/*=PC,  the  height  of  thefrus 

turn. 
A=area  of  the  greater  base, 
ffczrarea  of  the  less. 

AP— oC  :  PC  : :  aC  :  CV,  by  similar   trian-les  ; 
D        d  d       hd  a 

viz. —   -   —  :  h  i :  —  : =CV. 

2  2     D— d 

hd       AD 

but  PC+CV—Zi-f- -r =PV 

D— rf  D— d 
hd        a 

X — = solidity  of  the  pyramid  abV 

l)—d     3 
/*— D     A 


'■11 


X — zzsolidity  of  the  pyramid  ABV. 


D— d     3 
JiD       A        hd        a     AD— ad.    h 

Hence X X--= X— the  solidi- 

D— d    3       D— d     3         D— d     3 


i«)i>  Mensuration  of  Solids. 

(y  of  the  frustum  ABab.     Now  all  similar  figures  are 

h  other  as  the  squares  of  their  homologous  sides. 

(Euclid  VI,   ami  19th  ami  20th  ;  also  XII.  and  2d.) 

A      a 
Therefore  A  t  a : :  D*  :d*  or  — =r —  put  each  of  these 

equal  to  ?7i ;  then  A=7JiD2  and  «  — mcZ%  and  multiply- 
ing these  quantities  together,  we  get  aA  =  m2D2de,  con- 
sequently \/  aA=mY)d.  For  A  and  a  substitute  their 
values  in  the  expression  for  the  frustum's  solidity,  and  it 
7*il)3— mdth      m~- d*     k 

becomes X — = X — Xm=D2-f-Dc/-f  dx-, 

D — d       3       D— • d       3 

h  h 

X — xm=mD24-m*^+w^*  X  —  ;  restore  the  values, 

3  3  

of  A  and  a,  &c.  and  the  rule  becomes  A-j-v'aA-j-aj 

h 
X  — which  is  the  general  rule. 

3 

Corollary  1.  If  the  bases  be  squares,  of  which  D  and 

h 

d  are  each  a  side,  then  D*-fD</-{-eZ2  ;  X — is  the  soli- 

3 

dity  and  is  the  second  rule. 

Cor.  2.  If  the  bases  be  circles,  of  which  D  and  d 
are  diameters  then  D2  +  Drf-f-rf2  ;  X^X  .2618  is  the 
solidity,  which  is  the  third  rule. 

Cor.  3.  If  the  bases  be  regular  polygons,  and  t  rep- 
resent the  tabular  number  in  section  VIII.  also 
D  and  d  represent  a  side  of  each  base,  the  rule  be- 

h 

eomesD*x*+VD2*X^-H***  5  X— =D7-f Drf-f d»; 

3 
t 
X — xh,  the  solidity,  which  is  rule  4. 
3 


' 


Mensuration  of  Solids.  161 

That  the  rule  for  finding  the  superficies  of  a  frus- 
tum is  true,  will  readily  appear,  when  we  consider 
that  if  the  ends  he  regular  polygons,  the  sides  will 
consist  of  as  many  trapeziods  as  there  are  sides  in  the 
polygon  ;  and  the  common  height  of  these  trapeziods 
will  be  the  slant  height  of  the  frustum ;  and  the  sum  of 
their  parallel  sides,  the  sum  of  the  perimeters  of  the  ends* 

But  when  the  basis  are  circles,  the 
rule  is  not  so  obvious.  It  is  shewn  in 
section  VI.  that  the  curve  superficies  of 
a  right  cone  is  equal  to  the  rectangle  of 
half  the  circumference  of  the  base,  into 
the  length  of  the  slant  side :  viz.  that  it 
is  equal  to  the  area  of  the  sector  of  a 
circle,  whose  arch  is  equal  to  the  circum-  y 

ference  of  the  base,  and  radius  equal  to  ^ 

the  slant  side  of  the  cone. 

Let  DE=P,  the   circumference   of  the   great  hasc, 
and  BC— />,  the  circumference  of  the  less  base  of  the 
frustum  ;  AD  the  slant  height=S,  now,  because  simi- 
lar arches  of  circles  are  as  their  radii. 
DE  :  BC  : :  AD  :  AB 
S/> 
viz.  P    :   p     :  :     S    :   — 

P 

Sjp      P— .p 

Hence  B  D=A  D— A  B=S = xS 

P  P 

PS 
The  area  of  the  sector  A  D  E= — ,  and 

2 
P      Sp      Sp* 
The  area  of  sector  ABC= — x —  =  — 

2        P        2P 
Also  ADE — ABC = curve  surface    of  the  frustum, 
PS  8p*    PST— ®*S     P*_ p*  V+p 

BDEC= = = X  &= — : 

2      2Y  2P  2V  % 

V-p 

X- xS>  which  is  the  rule. 

P 

v2 


ifvj  Memura\  -  UiL<. 

Note.  From  the  foregoing  demonstration  we  have  the 
following  rule  for  finding  the  area  of  a  segment  of  a 
sector  B  1)  E  C,  or  the  front  of  a  circular  arch,  built 
with  stones  of  equal  length. 


KUI.E. 


Multiply  half  the  sum  of  the  bounding  arches  (RE 
and  BC)  by  their  distance  (BD),  and  the  product  will 
£?ive  the  area. 


Example  4.  If  each  side  of  the  greater  end  of  a 
piece  of  squared  timber  be  25  inches,  each  side  of  the 
less  end  9  inches,  and  the  length  20  feet ;  how  many 
solid  feet  are  contained  in  it  ? 

Jins.  By  rule  2,  43.1018  feet. 

5.  If  a  piece  of  timber  be  32  inches  broad,  and  20 
inches  deep,  at  the  greater  end;  and  10  inches  broad, 
and  G  inches  deep,  at  the  less  end ;  how  many  solid 
Feet  are  contained  therein,  the  length  being  18  feet? 

Jins.  By  the  general  rule,  37.3316  feet. 

6.  A  portico  is  supported  by  four  pillars  of  marble, 
each  haying  eight  equal  sides,  whose  breadth  at  the 
greater  end  is  7.5  inches,  and  at  the  less  end  1*  inches, 
and  their  length  12  feet  9  inches;  I  desire  to  know 
how  many  solid  feet  they  contain,  and  what  they  will 
come  to  at  12s.  lOd.  per  solid  foot  ? 

jns    C  Solidity  60.52927823  feet. 
I  Expence  I.  3S  :  16s.  :  9}d. 

7.  In  a  frustum  of  a  square  pyramid,  each  side  of 
the  greater  end  is  5  feet,  each  side  of  the  less  end 
3  feet,  and  the  height  8  feet;  required  the  height  and 
solidity  of  that  pyramid  of  which  this  frustum  is  a 
part  ?  n      5  TJie  height  is  20  feet,  see  p.  143. 

JlnS'  I  The  solidity  166$  feet. 

8.  If  each  side  of  the  greater  base  of  the  frustum  of 
a  hexagonal  pyramid  be  13,  each  side  of  the  less  base 
S,  and  the  length  24,  what  is  the  solidity  and  superfi- 
cies ?  '^   C  Solidity  Vo04.412S96. 

*\  Superficies  2141.7G42, 


Mensuration  of  Solids.  103 

§  VIII.  Of  the  Frustum  of  a  Cone. 

A  frustum  of  a  cone,  is  that  part  which  remains  when 
the  top  end  is  cut  oft*  by  a  plane  parallel  to  the  base. 

The  solidity  may  be  found  either  by  the  general  rule, 
or  rule  8,  of  the  preceding  section.  The  superficies 
may  also  be  found  by  the  rule  given  in  that  section. 

EXAMPLES. 


1.  Let  ABCD  be  the  frus- 
tum of  a  cone,  whose  greater 
diffmeter  CD  is  IS  inches, 
the  less  diameter  AB  9  in- 
ches, and  the  length  14.25 
feei ;  the  solid  content  is  re- 
quired. 


BY  THE  GENERAL  RULE. 

The  square  of  flp  is  324, 
which  multiplied  hy  .7854 
gives  254.4696  inches,  area 
of  the  greater  base.  The 
square  of  9  is  81,  which 
multiplied  by  .7854  gives 
63.6174,  area  of  the  less 
hasc. 


The  square   root   of    the 
16188.69433104  is  127.23AS. 
Area  of  the  greater  base 
Area  of  the  less  base, 
Mean  proportional 


product    of  these    areas 

254.4696 
-  '       -•        -         63.6174 

-  127.2348 


Sum  445.3218 
This    multiplied  by  4.75,  one  third  of  the    height, 
gives  2115.27855,  whieh^divide  by  144,  and  the  quo 
tient  is  14.68943  feet,  the  solidity. 


16*  Jhnsuraiion  of  Solids. 

Note.  The  diameter  of  the  less  base  being  exact- 
ly half  that  (if  the  greater,  in  this  example',  the  ope- 
ration might  have  been  performed  shorter:  for  the  area 
of  the  h'^s  base  in  sueh  eases  is  one  fourth  of  that  of 
the  greater, and  the  mean  area  double  that  of  the  less 
base,  or  half  that  of  the  greater. 

OR    THUS,    BY    RULE    3. 

The  square  of  18  (the  greater  diameter)  is  324,  and 
the  square  of  9  (the  less  diameter)  is  81,  and  the  rect- 
angle, or  the  product  of  £8  by  9,  is  162,  the  sum  of 
these  three  is  567,  which  multiplied  by  the  height 
11.25  feet,  gives  8079075  ;  which  multiplied  by  .2618, 
and  divided  by  144,  gives  14.68943  feet  as  before. 

To  find  the  Superficial  Content. 
The  circumference  of  the  greater  base,  by 

chap.  I.  section  IX.  is  56.5488 

And  the  circumference  of  the  les^s  base  is     28.2744 

Sum  84.8232 


Half  sum  42.4110 


To  the  square  of  4.5,  the  difference  between  the  ra- 
dii of  the  two  bases,  add  the  square  of  171,  the  per- 
pendicular height  in  inches  ;  the  square  root  of  the 
sum  2)261.27,  is  171.0592  inches,  the  slant  height; 
which  multiplied  by  42.4116  produces  7254.894J6672, 
to  which  add  the  sum  of  tm\,areas  of  the  two  ends, 
and  the  whole  superficies  will  be  7572.98136672  inches, 
or  52.r>    feet. 

2.  If  a  piece  of  timber  be  9  inches  in  diameter  at 
the  less  end,  36  inches,  at  the  greater  end,  and  24  feet 
long ;  how  many  solid  feet  are  contained  therein  ? 

Jns.  74.2203  feet. 

3.  If  a  piece  of  timber  be  136  inches  in  circumfer- 


Mensuration  of  Solids.  165 

ence  at  one  end,   32  inches   at  the  other,  and  21  feet 
long  5  how  many  solid  feet  are  contained  therein  ? 

Jns.  95.34816  feet. 


§  IX.  Of  a  Wedge. 

A  wedge  is  a  solid,  having  a  right  angled  parallel- 
ogram for  its  base,  and  two  of  its  sides  meeting  in  an 
ed^e. 

To  find  the  Solidity. 

RULE. 

To  twice  the  length  of  the  base  add  the  length  of 
the  edge,  multiply  the  sum  by  the  breadth  of  the  base, 
and  that  produet  by  the  perpendicular  height  of  the 
wedge;  and  one -sixth  of  the  last  product  will  be  the 
solidity. 

EXAMPLES. 

1.  The  perpendicular  height  01,  of  a  wedge  is  24.S 
inches ;  the  length  CK  of  the  edge,  110  inches;  the 
length  AE  of  the  base  70,  and  its  breadth  AB  30  inch- 
es; what  is  the  solidity  ? 

70  length  of  the  base  AE. 


■3' 


7500 

140                   24.8  perp.  01. 
110  length  of  the  edge  CK.  

60000 

250  sum.        .       30000 

30  breadth  of  the  base.  15000 


7500  product.  6)186000.0 

31000         inches;  which 
divided  by  1728  gives  17.9398  feet,  the  solidify. 


'-to  Mensuration  of  Solnh 

2.  Required  the  solidity  of  a  wedge,  whose  altitude 
^s  11  inehes,  in  inches,  and  the  length  and 
breadth  of  its  base  32  and  4.8  inches  ? 

Jlns.  892J  inches. 

3.  Required  the  solidity  of  a  wedge,  whose  altitude 
is  J  feet  -i  inches,  length  of  the  edge  3  feet  6  inch- 
es, length  of  the  base  5  feet  4  inches,  and  breadth  9 
inches  ?  Jns.  1, 13  tut  J,  &c.  feet. 


DEMONSTRATION    OF    THE    RULE. 

Let  ABCDEF  represent  a 
wedge  :  now  when  the  length 
of  the  edge  CD  is  equal  to  the 
length  AE  of  the  base  ABF 
E,  the  wedge  is  evid  ntly 
equal  to  half  a  parallelopipe- 
dt»  ABGCHDEF,  having 
the  same  base  and  altitude  as 
the  wedge ;  and  this  will  al- 
ways be  true,  whether  the  end  ABC  of  the  wedge  be 
perpendicular  to  its  base  ABFE,  or  inclined  as  ABI, 
since  parallelopipedons  standing  on  the  same  base  and 
between  the  same  parallels  are  equal  to  each  other 
{Euclid,  XT.  and  29.)  But  when  the  edge  CD  of  the 
wedge  is  longer  or  shorter  than  the  base,  by  any  quanti- 
ty DK  or  CI,  it  is  evident  that  the  wedge  will  be  great- 
er or  less  than  the  half  parallelopipedon  aforesaid,  by 
a  pyramid  whose  base  is  EFD, or  ABC,  and  perpendic- 
ular altitude  DK  or  CI.  Let  the  length  AE  of  the  base 
of  the  wedge  be  represented  by  L,  and  the  breadth  AB 
by  B :  call  the  length  of  the  edge  I,  and  the  per- 
pendicular height  AC,  «C,  or  01,  h.     Then,  the  so- 

BL 

lidity  of  the  wedge  will  be  expressed  by v^  or 

B/i             Bhh    L  2 

*— XL= ,  when  the  edge  is  of  the  same  length 


Mensuration  of  Solids.  1G7 

as  the  base :  The  solidity  of  the  pyramid  ABCI  will 
Bh     L—l 

be  expressed  by X -when  the  length  ID  of  the 

2  3 

edge  is  less  than  that  of  the  base ;  and  the  solidity  of 

Bh     I— L 

the  pyramid  EFDK  will  be  expressed  by x ? 

2  3 

when  the  length  of  the  edge  GK  is  greater  than  that 
Bhh       Bh      l—L     2L+1 

of  the  base.  Hence, •-£ X -= X  Bh, 

2  2  3  6 

the  rule,  when  the  edge  is  longer  than  the  base,   and 
Bhh    Bh      L—l     2L+1 

5 X =- XBh,   the   rule   when  the 

2        2  3  6 

edge  is  shorter  than  the  base. 


§  X.     Of  a  Vrismqid. 

A  prismoid  is  a  solid  somewhat  resembling  a  prism ; 
its  bases  are  right  angled  parallelograms,  and  parallel 
to  each  other,  though  not  similar,  and  its  sides  four 
plain  trapezoidal  surfaces. 

To  find  the  Solidity. 

RULE. 

To  the  greater  length  add  half  the  less  length,  mul- 
tiply the  sum  by  the  breadth  of  the  greater  base,  and 
reserve  the  product. 

Then,  to  the  less  length,  add  half  the  greater  length, 
multiply  the  sum  by  the  breadth  of  the  less  base,  and 
add  this  product  to  the  other  product  reserved  j  multi- 


16S 


ration  of  Solids. 


ply  the  sum  by  a  third  part  of  the  height,  and  the 
product  is  the  solid  content. 


EXAMPLES. 

1.  Let  ABCDEFGft  be  a  prismoid  given,  the  length 
of  the  greater  base  AB  38  inches,  and  its  breadth  AC 
16  inches;  aud  the  length  of  the  less  base  EF  30 
inches,  and  its  breadth  12  inches,  and  the  height  6 
feet;  the  solid  content  is  required. 

To  the  greater  length  AB  .?8,  add  half  EF  the  less 
length  15,  the  sum  is  53  ;  which  multiplied  by  in,  the 
greater  breadth,  the  product  is  848 :  which  reserve. 

Again,  to  EF  30,  add  half  AB  10,  and  the  sum  is 
49;  which  multiplied  by  12  (the  less  breadth  EG)  the 
product  is  588;  to  which  add  848  (the  reserved  pro- 
duct,) and  the  sum  is  1436  ;  which  multiply  by  2,  (a 
third  part  of  the  height,)  and  the  product  is  2872 y 
divide  this  product  by  144,  and  the  quotient  is  19.94 
feet,  the  solid  content. 

2.  One  end  of  a  prismoid  is  a  square,  each  side  of 
which  is  13 ;  the  other  a  right  angled  parallelogram, 


Mensuration  of  Solids.  169 

whose  length  is  12  and  breadth  5;  what  is  the  solidity, 
the  perpendicular  height  being  20  ?  Jlns.  22&3^. 

3.  In  the  neighbourhood  of  Newcastle,  and  in  the 
county  of  Durham,  &c.  the  coals  are  carried  from  the 
mines  in  a  kind  of  waggon,  in  the  form  of  a  prismoid. 
The  length  at  the  topis  generally  about  6 feet  9^  inch- 
es, and  breadth  4  feet  7  inches  ;  at  the  bottom  the 
length  is  3  feet  5  inches,  and  breadth  2  feet 5]  inches; 
and  the  perpendicular  depth  3  feetfli  inches.  Requir- 
ed the  solidity. 

Jlns.  126340.59375  cubic  inches,  or  73.11376  feet. 

4.  How  many  gallons  of  water,  reckoning  282  cubic 
inches  to  a  gallon,  are  contained  in  a  canal  304  feet 
by  20  at  top,  300  feet  by  16  at  bottom,  and  5  feet  deep  ? 

Ans.  166590.6383  gallons. 


DEMONSTRATION   OF  THE  RULE. 


The  prismoid  is  evidently  compassed  of  two  wedges 
whose  bases  are  equal  to  the  bases  of  the  prismoid, 
and  their  height  equal  to  the  height  of  the  prismoid. 
Let  L  equal  the  length  of  the  greater  base  AB,  which 
of  course  will  be  the  length  of  the  edge  of  the  less 
wedge ;  I  equal  the  length  of  the  less  base  EF,  or  the 
edge  of  the  greater  wedge :  B  the  breadth  of  the  great- 
er base  AC,  b  the  breadth  of  the  less  base  EG,  and  h 


1~0  .  'ids. 

th<>  common  height     Then  the  solidity  of  the  wedge 
whose  base  is  ABDC  will  be  expressed  by 
2L+1  l  y 

XB//,  and  the  solidity  of  the  wedge  whose  base  is 

6 

2L+1 

EFIIG  by xbh  ;    henec    the   solidity  of   the 

6 

2Z4-L  2/+L 

whole    prismoid    will  be  XBA,  X  X 

6  6 


L+l                    L  h 

bh— XB;  -f/-| Xb;  X— ,  which  is  the  rule 

2                          2  3 
exactly. 

This  rule  is  demonstrated  section  II.  Prob.  XIV.  of 
Emerson's  Fluxions,  and  in  a  similar  manner  at  page 
179,  second  edition,  of  Simpson's  Fluxions ;  also  in 
Holiday's  Fluxions,  page  302. 

L+/  B+& 

Note.  If =M,   and  =?n,  the   above   rule 

2  2 

h 
becomes  BL-J-W-f4rM?n ;   X  — ,  that  is,  To  the  sum  of 

6 
the  areas  of  the  two  ends  add  four  times  the  area  of 
a  section  parallel  to,  and  equally   distant  from,  both 
ends;  this  last  sum   multiplied  into   one-sixth   of  the 
height,  will  give  the  solidity. 

It  is  shewn  in  proposition  III.  page  456,  part  IV.  of 
Dr.  Button's  mensuration,  quarto  edition,  that  this  last 
rule  is  true,  for  all  frustum's  whatever,  and  for  all  sol- 
ds  whose  parallel  sections  are  similar  figures.  And, 
Mr.  Moss,  in  his  Guaging,  page  175,  third   edition, 


Mensuration  of  Solids.  171 

says,  that  it  is  nearly  true,  let  the  form  of  the  solid  be 
what  it  will. 

The  following  rule  for  measuring  a  cylindroid  given 
by  Mr.  Hawney,  from  Mr.  Everard's  Gu aging,  has 
been  by  some  teachers  considered  as  erroneous,  others 
again  have  imagined  it  to  be  true.  The  general  rule, 
which  has  just  been  described,  will  most  certainly  give 
the  content  either  exactly,  or  as  near  as  possible,  pro- 
vided the  figure  of  the  middle  section  between  CD  and 
AB  (see  the  following  figure)  can  be  accurately  deter- 
mined. Half  the  sum  of  CD  and  AB,  in  the  follow- 
ing example,  is  85,  and  half  the  sum  of  AB  and  EF  is 
20,  the  two  diameters  of  the  middle  section  ;  therefore 
it  must  either  be  an  ellipsis,  or  a  curve  of  the  oval 
kind,  whose  area,  perhaps,  cannot  be  easily  determin- 
ed. If  the  section  be  an  ellipsis  (and  it  cannot  mate- 
rially differ  from  one)  the  rule  given  by  Mr.  Hawney 
exactly  agrees  with  the  above  general  rule.  But  Mr. 
Moss,  in  prob.  VI.  sect.  VIII.  of  his  Guaging,  has 
shewn  that  the  figure  of  the  middle  section  between 
CD  and  AB,  can  never  be  an  ellipsis,  unless  the  par- 
allel ends  AB  and  CD  are  similar  ellipsis,  and  simi- 
larly situated ;  viz.  the  transverse  and  conjugate  diam- 
eters of  each  end,  respectively  parallel  to  each  other, 
and  this  circumstance  can  never  happen  but  when 
the  solid  is  the  frustum  of  an  elliptical  cone.  The 
rule  therefore  cannot  be  strictly  true,  though  sufficient- 
ly near  for  any  practical  purpose. 

To  measure  a  Ctlindroid  ;  or,  the  Frustum  of  an 
elliptical  Cone. 

RULE. 

To  the  longer  diameter  of  the  greater  base,   add 
half  the  longer  diameter  of  the  legs  base,  multiply  the 


I7J 


Mensuration  of  Solids. 


sum  by  (he  shorter  diameter  of  the  greater  base,  and 
e  the  product 

Then,  to  the  longer  diameter  of  the  less  base,  add 
half  the  longer  diameter  of  the  greater  base;  and 
multiply  the  sum  by  the  shorter  diameter  of  the  less 
add  tliis  product  to  the  former  reserved  pro- 
duet,  and  multiply  the  sum  by  .2018  (on  -third  of 
)  and  then  by  the  height,  the  last  product  v  ill  be 
the  solidity. 

EXAMPLES. 


1 .  Let  ABCD  be  a  eylindroid,  whose  bottom  base  is 
an  ellipsis,  the  transverse  diameter  being  41  inches ; 
and  the  conjugate  diameter  14  inches;  and  the  upper 
base  is  a  circle,  of  which  the  diameter  is  26  inches ; 


Mensuration  of  Solids.  173 

and  the  height  of  the  frustum  is  9  feet ;  the  solidity 
is  required. 


44=  CD 

26=AB 

I3=half  AB 

22=half  CD 

— 

— 

57     sum 

48     sum 

14=EF 

26=AB 

228 

288 

57 

96 

798  product  reserved. 

1248   product.      * 

798  add  reserved  product. 

2046  sum. 

2046 

.2618 

16368 

2046 

12276 

4092 

533.6428 

9= height. 

4820.7852  which,   divided   by 
144,  gives  33.4776  feet  the  solidity. 

2.  The  transverse  diameter  of  the  greater  base  of  a 
cylindroid  is  13,  conjugate  8;  the  transverse  diameter 
of  the  less  base  10,  and  its  conjugate  5.2.  If  the 
length  of  the  cylindroid  be  20,  what  is  the  solidity  ? 

Ms.  1203.2328. 

3.  I  desire  to  know  what  quantity  of  water  an  ellip- 
tical bath  will  hold,  the  longer  diameter  .at  the  top  be- 

Q    2 


Mmsuration  of  Sulich. 


2  feet,  ami  shorter?;  (he  longer  diameter  at  the 
boll 0111  10  feet,  and  the  shorter  6  feet  ;  and  the  depth 
i  feet :  reckoning  2%2  eubic  inches  to 


a  gallon  ? 


Jlns.  1379.6303  gallons. 


§  XI.    Of  a  Sphere  or  Globe. 

A  sphere,  or  globe,  is  a  round  solid  body,  every  part 
of  its  surface  being  equally  distant  from  a  point  within, 
called  its  eentre;  and  it  may  be  conceived  to  be  form- 
ed by  the  revolution  of  a  semicircle  round  ils  diameter. 

To  find  the  Solidity. 

RULE    I. 

Multiply  the  cube  of  the  diameter  by  .5236,  and  the 
product  will  be  the  solidity. 

RULE    II. 

Multiply  the  diameter  of  the  sphere  into  its  circum- 
ference, and  the  product  will  be  the  superficies; 
which  multiplied  by  one-third  of  the  radius,  or  a  sixth 
part  of  the  diameter,  will  give  the  solidity. 


EXAMPLES. 


1.  Required  the  soli- 
dity of  a  globe,  ABCD,  / 
whose  diameter  AB   or  $.„ 
CD  is  20  inches.  \ 


Mensuration  of  Solids.  175 

BY  RULE  1. 

20X20X  20  X  .5236=4188.8,  the  solidity. 

BY  RULE  2. 

20X3.1416=62.832,  the  circumference. 
20X62.832=1256.64.  the  superficies. 
20xl256.64~6=41SS.S,    the    solidity,    in    inches, 
=  2.424  feet. 

2.  The  diameter  of  the  earth  is  7970  miles,  what  is 
its  surface  in  square  miles,  and  solidity  in  cubic  miles  ? 

a        (  Surface  199537259.44  miles. 
*i)lS'   I  Solidity  265078559622.8  miles. 

3.  The  circumference  of  a  sphere  is  1,  what  is  its 
solidity  and  superficies  ? 

a        S  Soli(lity  -0168868. 
"      *  \  Superficies  .3183099. 

4.  What  is  the  solid  and  superficial  content  of  a 
globe,  whose  diameter  is  30  ? 

-        J  Solidity  14137.2. 
J1VS'  I  Superficies  2827.44. 

5.  The  circumference  of  a  globe  is  50.3,  what  is  its 
solidity  and  superficies  ? 

^        S  Solidity  2149.073728. 
£  Superficies   805.3526, 

6.  A  globe,  ti  cube,  a  cylinder, 

All  three  in  surface  equal  are,*  viz.  3.1416 
In  solidity  what  do  they  ditt'er .?         each. 
Jlns.  The  solidity   of  the  globe  is   .5237,  the  cube 
.378877,  and  cylinder  .4275176  ;  so  that  of  all  solids, 
under  the  same  superficies,  the  globe  is  the  greatest. 

*  First,  for  the  globe.  If  the  superficies  of  a  globe  be 
divided  by  3.1416,  the  quotient  will  be  the  square  of 
the  diameter,  the  square  root  of  whieh  will  be  the 
diameter. — Hence  the  solidity  is  easily  found, 


170 


Mensuration. of  Solids. 


Second,  for  the  cube.  The  cube  lias  6  equal  surfaces.* 
therefore  the  whole  surface  divided  by  6?  will  give  the 
square  of  the  side  of  the  cube  ;  the  square  ruol  will 
be  the  hide. — Hence  the  solidity  is  found. 

Third,  for  the  cylinder.  The  diameter  and  depth  of 
the  cylinder  are  here  supposed  equal. — The  superficies 
of  a  cylinder,  whose  diameter  and  depth  are  each  an 
unit,  will  be  found  to  be  4.7124;  and  the  superficies  of 
similar  cylinders  are  to  each  other  as  the  squares  of 
their  diameters,  therefore  4.7124  is  to  the  square  of  1, 
as  the  superficies  of  the  cylinder  is  to  the  square  of 
its  diameter;  the  square  root  of  which  will  be  the 
diameter,  and  likewise  the  depth. — Hence  the  solidity 
of  the  cylinder  is  found. 


DEMONSTRATION  OF  THE  RULE,  &C. 

Tliat  every  Sphere  is  two-thirds  of  its  Circumscribing 
Cylinder,  may  be  thus  proved. 

Let  the  square  ABCD, 
the  quardrant  CBD,  and 
the  right  angled  triangle 
ABD,  be  supposed  all 
three  to  revolve  round 
the  line  BI)  as  an  axis  : 
Then  will  the  square 
generate  a  cylinder,  the 
quadrant  a  hemisphere, 
and  the  triangle  a  cone, 
all  of  the  same  base  and 
altitude. 

By  Euclid,  I.  and  47,  FD*--FH*-fDH%  but  FIX 
EH,  and  GHrrDH  because  EH  is  parallel  to  CD, 

therefore  EH*=FHs4-GHa ;  and  as  circles  are  to 
each  other  as  the  squares  of  their  diameters,  or  radii 
(Euclid,  XII.  and  2,)  it  follows  that  the  circle  describ- 
ed by  EH,  is  equal  to  the  two  circles  described  by  FH 


Mensuration  of  Solids.  177 

and  GH ;  take  away  the  circle  described  by  FH,  and 
there  remains  the  circle  described  by  EH— -circle  de- 
scribed by  FH  (viz.  the  annul  us,  or  ring,  described  by 
EF  between  the  sphere  and  cylinder)— circle  describ- 
ed by  GH.  And  it  is  evident,  that  this  property  will 
hold  in  every  section  similar  to  EH ;  but  the  cone, 
hemisphere,  and  cylinder,  may  be  conceived  to  be  made 
up  of  an  infinite  number  of  such  sections,  therefore 
when  the  hemisphere  is  taken  out  of  the  cylinder,  the 
remaining  part  is  equal  to  the  solidity  of  the  cone ; 
but  the  cone  is  one-third  of  the  cylinder,  therefore  the 
hemisphere  must  be  two-thirds,  and  what  is  here 
proved  with  respect  to  the  halves  of  the  proposed  so- 
lids, holds  equally  true  with  the  wholes.  Therefore 
every  sphere  is  two-thirds  of  its  circumscribing  cylin- 
der. Now  if  D  be  the  diameter  and  height  of  a  cylin- 
der, its  solidity  will  be  D*x.~S54xDzrD3X.78.34, 
hence  the  solidity  of  its  inscribed  sphere  will  be  D3X 
.7S3-1X|=D3  X^236,  which  is  the  first  rule. 

'The  Surface  of  a  Sphere  is  equal  to  the  Curve  Surface 
of  its  Circumscribing  Cylinder, 

Take  FK,  an  extremely  small  part  of  the  quadran- 
tal  arch  CB,  and  suppose  lines  LM  and  EH  to  be 
drawn  through  K  and  F  parallel  to  each  other.  Then, 
because  FK  is  extremely  small,  it  may  be  considered 
as  a  straight  line,  and  the  angle  FKD  as  a  right  an- 
gle. The  figure  FKMH,  by  revolving  round  MH, 
will  form  the  frustum  of  a  cone,  whose  slant  side  is 
FK ;  and  the  figure  ELMH,  by  revolving  in  the  same 
manner,  will  form  a  cylinder,  whose  perpendicular 
height  is  LE.  But  the  surface  of  the  frustum  is 
equal  to  FKXhalf  the  sum  of  the  circumferences  of 
two  circles  whose  radii  are  KM  and  FH ;  or  these  be- 
ing extremely  near  to  each  other,  the  surfaee  of  the 
frustum  will  be  equal  to  FKX  circumference  of  a  cir- 
ele,  whose  radius  is  KM,  or  FH.  Let  C— the  circum- 


178  Mensuration  of  Solids. 

ferenee  of  a  circle  whose  radius  is  (  D  er  EH,  ami 
c— the  circumference  of  a  circle  whose  radius  is  FH 
or  KM:  then  the  surface  formed  bj  the  revolution  of 
FkMH  about  Mil,  is  equal  to  cXFKj  ami  the  cylin- 
drical surface  formed  by  the  revolution  of  ELMH,  is 
equal  to  CxLE.  But.  the  triangles  FeK  and  DMK 
are  equiangular  and  similar:  for  FKD  and  eKM  are 
right  angles;  take  away  the  common  angle  eKG,  and 
there  remains  FKc— DKM;  and  the  angles  FeK  and 
KMD  are  right  angles  :  therefore 

DK  or  CD  :  KM  :  :  KF  :  Ke  or  LB. 
But  the  circumferences  of  circles  are  to  each  other  as 
their  radii,  therefore 

C  -.  c  :  :  CD  or  LM  :  KM,  consequently. 
C  :  c  :  :  KF  :  Ke  or  LE. 
Hence  CxLE=cXKF.  That  is  the  cylindrical  sur- 
face formed  by  the  revolution  of  ELMH,  is  equal  to 
the  spherical  surface  formed  by  the  revolution  of 
FKMH.  And,  if  more  parallel  planes  be  drawn,  ex- 
tremely near  to  each  other,  the  small  parts  of  the 
cylindrical  surface  will  be  equal  to  the  correspondent 
parts  of  the  spherical  surface  :  and  therefore  the  sum 
of  all  the  parts  of  the  cylindrical  surface,  will  be 
equal  to  the  sum  of  all  the  parts  of  the  spherical  sur- 
face; that  is,  the  surface  of  the  half  cylinder  will  be 
equal  to  the  surface  of  its  inscribed  hemisphere,  and 
the  surface  of  the  whole  cylinder  equal  to  the  surface 
of  its  inscribed  sphere.  Hence  if  D=^the  diameter  as 
before,  then 

D 

DX3.1416XH;  X— is  the  second  rule,= 

6 

D3X5236,  the  same  as  the  first  rule. 

Corollary  1.  The  surface  of  the  sphere  is  |  of  the 
whole  surface  of  its  circumscribing  cylinder. 

For  D*X3. 1416  is  the  surface  of  the  sphere=rcMrre 
surface  of  the  cylinder ;  and  the  area  of  the  two  ends 


-Mensuration  of  Solids. 


179 


ofthecyliiKler=D2X.7S54;+D2X.7854:=l,570SD% 
therefore  the  surface  of  the  whole  cylinder  is=3.1416 
D2 -|- 1.5708  D2 =4.7124  D2  j  two-thirds  of  which = 
3.1416  D2,  the  surface  of  the  sphere. 

Cor.  2.  Hence  the  surface  of  the  whole  sphere  is 
equal  to  the  area  of  four  great  circles  of  the  same 
sphere;  or  to  the  rectangle  of  the  circumference  and 
diameter. 

Cor.  3.  The  surface  of  any  segment  or  zone  of  a 
sphere,  is  equal  to  the  curve  surface  of  a  cylinder,  of 
the  same  height;  and  whose  diameter  is  equal  to  that 
of  the  sphere. 

For  the  zone  formed  by  the  revolution  of  FKMH 
about  MH,  is  equal  to  the  surface  of  the  cylinder  form- 
ed by  the  revolution  of  ELMH. 

Investigation  of  General  Rules  for  finding  the  Solidity 
of  any  Segment,  or  Zone  of  a  Sphere. 

Let  ACy  represent  a  tri- 
angular pyramid,  whose  side 
Aw  is  infinitely  small.  The 
sphere  may  be  considered  to 
he  constituted  of  an  infinite  P, 
number  of  such  pyramids,  Jj 
yCe,  eCrt,  &c.  whos^  bases  *1 
compose  the  spherical  sur- 
face, and  altitudes  are  equal  « 
to  the  radius  of  the  sphere, 
their  common  vertex  being 
the  centre  C.  And,  conse- 
quently, the  sphere,  or  any  sector  thereof,  is  equal  to  a 
pyramid,  the  area  of  whose  base  is  equal  to  the  spher- 
ical surface,  and  height  equal  to  the  radius  of  the 
sphere.  Let  6T  the  height  of  the  segment  aTm  be 
called  h,  the  diameter  of  the  sphere  D,  then  will  its 
circumference  be  3.1416  D;  hence  by  Cor.  3,  prece- 
ding, the  surface  of  the  spherical  segment  aTm  will  be 
expressed  by  3.1116  Bh.  And  the  solidity  of  the  spher- 


I  so  Mensuration  of  Solids. 

ical    sector   CaTm  (supposing  CM  to   be  drawn)  by 

what  has  just  been  observed,  will  be  3.1410  D/*X — = 

6 

,5230  D2//. 

By  the   property    of  the  circle  D — hXh—ab2  ;  but 

bU    '  D—2h   

«6'X3.1416X— =D— /iX/iX3.1416X =D-WtX/t 

3  6 

X.5236X13— 2/l  =  DT/i— 3D/l84-2/i3;X.3236,  the  so- 
lidity of  the  cone  Cam.  Now,  the  solidity  of  the  cone 
taken  from  the  solidity  of  the  sector,  leaves  the  solidity 
of  the  segment,  viz. 

D "  h  X  .523  6  5— -D  *  /t-f3D/i '  —2h  3  ;  X  .5236  = 


3D/tT- -2/i3  ;  X  .5236  =  3D— 2hXh'*X  .5236,  which  is 
one  rule  for  the  solidity  of  a  segment  ;  and,  it  is  evi- 
dent from  the  property  of  the  circle,  that  this  rule 
will  be  true,  whether  the  segment  be  greater  or  less  than 
the  hemisphere,  or  whatever  be  the  magnitude  of  h, 
provided  it  be  not  greater  than  D.     Or,  if  r—ab  the  ra- 


dius of  the  segment's  base,  then  D — hxh=r*,  there* 
r€xhJ 

fore  D= ,  for  D  in    the  above  rule,  substitute 

h  8r*X3&* 

its  value,  and  the  rule  becomes 2h;  xh*  X 

— h 

.5236  =  3r9-j-/£2x/iX.5230,an  useful  rule,  when  the  dia- 
meter of  the  sphere  is  not  given. 

To  find  the  Solidity  of  a  Zone  of  a  Sphere. 

Call  the  height  of  the  greater  segment  H,  and  the 
height  of  the  less/* ;  also  11  the  radius  of  the  greater 
base,  and  r  that  of  the  less :  then  it  is  evident  the  dif- 
ference between  the    solidity   of  these  two    segments 


will  be  the  solidity  of  the  zone.    Hence  3R*-fHTxH 


Mensuration  of  Solids.  jsi 


X.5236;-3r24-/i2X/iX.^36  =  (3R2H4.H3;-3^Vt+/^) 
X.5236.  Put  a—H—h  the  breadth  of  the  zone, 
and  D  the  diameter  of  the  sphere  :  then,  by  the  prop- 
erty of  the  circle,  D—H  xH=R2,and  D— hxh=r* 

ItXH2 

from  the  former  of  these  D= ,  and  from  thelat- 

H 
r2+/i2  R8-fH*     r2-f/*2 

ter  D= ~;  therefore = ,  and    from 

h  H  h 

the  above  a=-H — h.     Exterminate  the  values  of  H  and 


h,  and  the  above  Theorem  (3R2H+HT;— -3r2/i-f-/i3)  : 

«2 

X.5236,  will  become R2-fr2-j ;XaX1.5708. 

3 
If  one  of  the  radii  pass  through  the  centre,  as  in  the 
I)2     —    — 
zone   A  G  m  a,    then  R2=— =r«&2-H>C*=r2-f  a* ; 

4 
henee  the  last  Iheorem  becomes  r2-j-|a2 ;  X«X  3.1416 
=  -lD* — !-«*;  XaX3.1416. 

Hence  rT-f  fa2 :  X«X6.2832=:iD2 — 1«2X«X6.2S32 
will  express  the  solidity  of  the  middle  zone  amKN, 
being  double  of  the  former,  where  a  is  half  the  alti- 
tude, and  r=half  the  diameter  of  each  end.  Put  A 
for  the  whole  altitude,  and  d=2r  the  diameter  of  each 
end,  and  the  theorems  become  rf'+fA* ;  XAX  .7854 
— D? — |A2~;  X AX- 7854. 

To  find  the  Solidity  of  the  Segment  of  a  Sphere. 

RULE    I. 

From  three  times  the  diameter  of  the  sphere  sub- 
tract twice  the  height  of  tfie  segment  ;  multiply  the 
remainder  by  the  square  of  the  height,  and  that  pro- 
duct by  .5236,  and  the  last  product  will  be  the  solidity. 

.  R 


is  j  Mmisuratkm  of  Solids. 


Rt'LE    I! 


To  three  times  the  square  of  the  radius  of  tli- 
incut's  base,   add    (lit-  square  of  its   height  ;  multiply 
thii  sum  by  the  height,  and  the  product  by  .5236,  and 
(he  last  product  will  be  the  solidity. 


EXAMPLES. 


1.  Let  ABCD  be  the  frustum   a^^T^Vti 
of  a  sphere  ;  suppose    AB,  the    /  ti        "   \ 

diameter  of    the  frustum's  base  /  \ 

to  be  16  inches,  and  CI),  the  I 
height,  4  inches  ;  the  solidity  is  I 
required  ?  V 


BY    RILE 


Al)*-j-lX\.f!)C=CE;  that  is,  8X8—4,4-4=20 
the  diameter  ;  therefore,  (20X3 — 4X2)X4X^  *  .5236 
=  433.6352,  the  solidity  of  the  frustum. 

by  rule  2. 

(  3AD2+CD2)XCDX.3236=:435.6352,  the  solidi^ 
ty,  as  before. 

2.  What  is  the  solidity  of  a  segment  of  a  sphere,- 
whose  height  is  9,  and  the  diameter  of  its  base  20? 

Ms.  1795.4214. 

3.  What  is  the  solidity  of  the  segment  of  a  sphere, 
whose  diameter  is  20  feet,  and  the  height  of  the  seg- 
ment 5  feet  ?  Ms.  654.5  feet. 

4.  The  diameter  of  a  sphere  is  21,  what  is  the  so- 
lidity of  a  segment  thereof,  whose  height  is  4.5  ? 

Ms.  572.5366^i 

5.  The  diameter  of  the  base  of  a  segment  of  *y 
sphere  is  2$,  and  the  height  of  the  segment  6.5 ;  re- 
quired the  solidity  ?  Jins.  2144.99283. 


Mensuration  of  Solids.  183 

To  find  the  Solidity  of  the  Frustum,  or  Zone,  of  a  Sphere, 

GENERAL    RULE. 

Add  together  the  square  of  the  radii  of  the  ends, 
and  one-third  of  the  square  of  their  distance,  viz.  the 
lieight;  multiply  the  sum  by  the  height,  and  that  pro- 
duct by  1.5708,  the  last  product  will  be  the  solidity. 

Or,  for  the  middle  Zone  of  a  Sphere. 

i  To  the  square  of  the  diameter  of  the  end,  add  two- 
thirds  of  the  square  of  the  height;  multiply  this  sum 
by  the  height,  and  then  by  .7854,  for  the  solidity. 
f  2?  Or,  from  the  square  of  the  diameter  of  the  sphere, 
subtract  one-third  of  the  square  of  the  height  of  the 
middle  zone;  multiply  the  remainder  by  the  height, 
and  then  by  .7854,  for  the  solidity. 

EXAMPLES. 

1.  Required  the  solidity  of  the  zone  of  a  sphere 
AGKN  (see  the  figure,  p.  179,)  whose  greater  diame- 
ter AG  is  20  inches,  less  diameter  NK  15  inches,  and 
distance  between  the  ends,  or  height  CB  10  inches. 

(AC2+NB2+1CB2)XCBX1.5708=2977.9741764, 
the  solidity. 

2.  Required  the  solidify  of  the  middle  zone  am  KN 
of  a  sphere,  whose  diameter  AG  is  22  inches ;  the  top 
and  bottom  diameters  a  m  and  NK  of  the  zone  being 
each  16.971  iuches,  and  the  height  B6  14  inches  ? 

Jlns.  4603.4912  inches, 

3.  Required  the  solidity  of  a  zone  whose  greater 
diameter  is  12,  less  diameter  10,  and  height  2  ? 

Alls.  195.8264. 

4.  Required  the  solidity  of  the  middle  zone  of  a 
sphere,  whose  top  and  bottom  diameters  are  each  3 
feet,  and  the  breadth  of  the  zone  4  feet  ? 

Ms.  61.7848  feet. 


is*  Mensural  ulids. 

5.  Required  the  solidity  of  the  middle  zone  of"  a 
sphere,  whose  diameter  is  5  feet,  and  the  height  of 
the  zone  4  feet  ?  Jfiw.  61.7848  feet. 

To  find  the  Convex  Surface  nf  any  Segment,  or  Zone, 

of  a  Sphere. 

RULE. 

Multiply  the  circumference  of  the  whole  sphere  by 
the  height  of  the  segment,  or  zone,  and  the  product 
will  be  the  convex  surface. 

EXAMTLES. 

1.  If  the  diameter  of  the  earth  be  7970  miles,  the 
height  of  the  frigid  zone  will  be  552.361283  miles  ; 
required  its  surface  ? 

Ms.  6318761.10718.2210  miles. 

2.  If  the  diameter  of  the  earth  be  7970  miles,  the 
height  of  the  temperate  zone  will  be  2143.6235335 
miles ;  required  its  surface  ? 

Ms,  53673229.812734532  miles. 

3.  If  the  diameter  of  the  earth  be  7970  miles,  the 
height  of  the  torrid  zone  will  be  3178.030327  miles  ; 
required  its  surface  ? 

Ms.  79573277.600166501. 


Note.     The  surfaces  of  the  . 

POT zones'  b-y 


} 

The  surfaces  of  the  twolemO  B367322c,.81273i332 

perate  zones,  by  example  2,  I  03a73229m273is3i 

are J  * 

The  torrid  zone,  example  3,5  79573277.600166504 


is 


Surface   of  the   whole  globe,! 

agreeing  exactly   with  ex-  I   199557259.440000000 
ample  2,  p.  175.  -     -     -     J 


Mensuration  of  Solids. 


185 


§  XII.     Of  a  Spheroid. 

A  spheroid  is  a  solid  formed  by  the  revolution  of 
an  ellipsis  about  its  axis.  If  the  revolution  be  made 
about  the  longer  axis,  the  solid  is  called  an  oblong,  or 
prolate  spheroid,  and  resembles  an  egg :  but  if  the 
revolution  be  made  round  the  shorter  axis,  the  solid  is 
called  an  oblate  spheroid,  and  resembles  a  turnip,  or 
an  orange.  The  earth  aud  all  the  planets  are  consi- 
dered as  oblate  spheroids.  In  an  oblong  or  prolate 
spheroid,  the  shorter  axis  is  called  the  revolving  axis, 
and  the  longer  axis  the  fixed  axis :  but  in  an  oblate 
spheroid,  the  longer  axis  is  called  the  revolving  axis, 
and  the  shorter  axis  is  called  the  fixed  axis. 

To  find  the  Solidity. 


Multiply  the  square   of  the  revolving   axis  by  the 
fixed  axis,  and  that  product  by  .5236  for  the  solidity. 

EXAMPLES. 


1.  Let  ADBC  be  a  prolate  sphe- 
roid, whose  longer  axis  CD  is  55 
.inches,  and  shorter  axis  AB  33 ;  re- 
quired the  solidity  ? 

AB*XCDx  .5236=31361.022,  the 
solid  content,  required. 


2.  What  is  the  solidity  of  an  oblate  spheroid,  whose 
longer  axis  is  .55  inches,  and   shorter  axis  33  inches  ? 

Ans.  52263.37  inches. 
r  2 


ISC, 


•WaisaraiioH  of  Solids. 


3.  Th<>  axes   of  an  oblong  or  prolate  spheroid  are 
SO  ami  30;  required  the  solidity?  Jlns.  3S06& 

4.  The  axes   of  an  oblate  spheroid  are  50  and  30 ; 

required  the  solidity?  Ms.  392,0 

re 


Demonstration.  Suppose  the  figure  NTwSN  to 
represent  a  spheroid,  formed  by  the  rotation  of  the 
semi-ellipsis  TNS,  about  its  transverse  axis  TS. 

Let  D=TS,  the  length  of  the  spheroid,  and  the 
axis  of  the  circumscribing  sphere  ;  and  rf=Nn,  the 
diameter  of  the  greatest  circle  of  the  spheroid  : 


Then  TS*  :  Nw»   ::  A6*  :  ab\  by  section  XV.  p. 


127,  that  is,  D2  :  d*  ::  Aft2  :  a&2 ;  but  all  circles  are 
to  each  other  as  the  squares  of  their  radii.  Now,  the 
solidity  of  the  sphere  may  he  imagined  to  be  constitut- 
ed of  an  infinite  numb  r  of  circles,  whose  radii  are 
parallel  to  Aft  ,•  and  the  spheroid  of  an  infinite  number 
of  circles  whose  radii  are  parallel  to  ab.  Therefore 
D2  :  d-  ::  solidity  of  the  sphere  whose  diameter  is  D  : 
Solidity   of  the   spheroid  whose   revolving  axis   is  d; 

3230  D3  d2 

viz.    D2   :  &*   ::  D3X.5236  : =.5236  Bdz. 

D2 
Again  for  the  oblate  spheroid,  by  Emerson's  Conies, 


Book  I.  prob.  7,  NC2  :  TC2  ::  naXaS  :  Ba° ;  but  by 


Mensuration  of  Solids.  1S7 

the  property  of  the  circle  naXfflN=6d% 

therefore  NO  :  TC2  : :  6a2  :  Ba2,  viz. 

d2  :  D2  ::  ba*  :  Ba2;  hence  from  the 
same  manner  of  reasoning  as  above,  d2  :  D2  : :  solid- 
ity of  *he  sphere,  whose  diameter  is  d  :  the  solidity  of 
a  spheroid,  whose  revolving  axis  is  D. 

.523Qd3&2 

Hence  d2  :  D2  ::  d3x-5236  : =.52SQdD7-. 

d2 
Now  from  these  proportions,  between  the  sphere  and 
its  inscribed  or  circumscribed  spheroid,  it  will  be  very 
easy  to  deduce  theorems  for  finding  the  solid  content, 
either  of  the  segment  or  middle  zone  of  any  spheroid, 
having  the  same  height  with  that  of  the  sphere ;  for, 

As  the  solidity  of  the  whole  sphere  is  to  the  solidity 
of  the  whole  spheroid,  so  is  any  part  of  the  sphere  to* 
the  like  part  of  the  spheroid. 

1st.  To  find  the  solidity  of  a  segment  of  a  spheroid. 
TSa  :Nrc2  ::  Ab*  :  ab2  in  the  prolate  spheroid, 
Nw2:TS2  ::  ab2  :  aB2  in  the  oblate  spheroid, 

Hence       TS2  :  Nn2  : :  solidity  ATM :  solidity  aYm, 

Nw3  :  TS2  ::  solidity  bNb  :  solidity  BNB. 
But  the  solidity  of  ATM  or  6N6  may  be  found  by  rule 
the  first,  section  XL  p.  181. 

2d.  To  find  the  solidity  of  the  middle  zone  of  a  spheroid, 
Let/=TS  the  fixed  axis.  }    for  the  oblong 

r=Nn  the  revolving  axis      $        spheroid. 
h  —  Bb  the  height  of  the  middle  zone  or  frustum. 
D=AM  the  diameter  of  one  end  of  the  spherical 

zone. 
d=am  the  corresponding  diameter  of  the  spheroid- 
al zone. 


184  Mensuration  of  Solids. 

By  section  XI.    the  solidity   of  the    middle   zone   of 

the  sphere  is    D8-j X//X  .7854=3l)*-f  2/t«  ;  X^X 


D2 


.2618  ;   hut   A6*  =  AC2— Co*    vfcfc.  --  s& ; 

4  4  4 

therefore  J)2=/a — //2,  hence  the  solidity  of  the  spher- 
ical zonc=3/2 — h2  :  xhX-^iH.  JB.ut.5236/3  :  ..1236 
frr  ::  Zf2—h;  Xhx. 2(518  :  the  solidity  of  the  spher- 
oidal frustum,  vis. 

/■  :  ?•::  3]f  *•— A*  $  X/*X  .2618  :  the  solidity  of  the 
spheroidal  frustum.     By  tlie  property  of  the  ellipsis 

I\SB  :  N?*2  : :  AM2  :  am2,  that  is, 

r2h2 

f2  :  r'  ::f2—h2~  :  rf2?  hence /2  = $ 

r2— </2 
r2/i2  2r2h 

Consequent.  :  r2  : : h2  ;  X^X-2618: 

r2 — f/z  r2 — rf2 

the  solidity  of  the  spheroidal  frustum =2r2-f-^/2  : 
Xh  X  .2618.  The  rule  for  the  oblate  spheroidal  zone 
will  be  exactly  the  same,  putting  r  for  TS  and  d  for 
BoB. 

To  find  the  Solidity  of  the  Segment  of  a  Spheroid  ;  the 
Base  of  the  Segment  being  parallel  to  the  revolving 
Axis  of  the  spheriod. 


From  three  times  the  fixed  axis  subtract  twice  the 
height  of  the  segment,  multiply  the  remainder  by  the 
square  of  the  height,  and  that  product  by  .5236. — 
Then  as  the  square  of  the  fixed  axis,  is  to  the  square 
of  the  revolving  axis ;  so  is  the  last  product,  found 
above,  to  the  solidity  of  the  spheroidal  segment. 


Mensuration  of  Solids.  139 


EXAMPLES. 


1.  In  &  prolate  spheroid,  the  transverse  or  fixed  axis 
TS  is  100,  the  conjugate  or  revolving  axis  7$n  is  60, 
and  the  height  //T  of  the  segment  aTm  is  10 :  required 
the  solidity  ? 

(3TS—2T&)  XWX. 5236= 14660.8  ;  then  as 
TS*  :  Nn2  :  :  1-1660.8  :  5277.888,  the  solidity  of  the 
segment  aTm. 

2.  In  an  oblate  spheroid,  the  transverse  or  revolving 
axis  TS  is  100,  the  conjugate  or  fixed  axis  Nn  is  60, 
and  the  heigh:  Na  of  the  segment  BNB  is  10:  requir- 
ed the  solidity  ? 

(3.Nn— 2Na)  XN«2X.5236=:8377.6  ;  then,  as 
Nn*  :  TS*  ::  8377.6  :  2327 1£,  the  solidity  of  the  seg- 
ment BNB. 

Now,  square  of  Nn=3600  and  square  of  TS  — 10000. 
As  3600  : 4  0000  ::  8377.6  :  23271-L  solidity  of  the  spher- 
oidal segment  BNB. 

3.  The  axes  of  a  prolate  spheroid  are  50  and  30 ; 
what  is  the  solidity  of  a  segment  whose  height  is  5, 
and  its  base  parallel  to  the  conjugate  or  revolving  ax- 
is ?  Ans.  659.736./'  , 

4>.  The  axes  of  an  oblate  spheroid   are  50  and  3o/j~S 
what  is  the  solidity  of  a  segment   whose  height   is  6,      J 
and  its  base  parallel  to  the  transverse  or  revolving  ax- 
is ?  Jlns.  4084.08. 

5.  If  the  axes  of  a  prolate  spheroid  be  10  and  6,  re- 
quired the  area  of  a  segment  whose  height  is  1,  and 
its  base  parallel  to  the  conjugate  or  revolving  axis  ? 

Jlns.  5.27788S. 

6.  The  axes  of  a  prolate  spheroid  are  36*  and^.8; 
what  is  the  solidity  of  a  segment  whose  height 
and  its  base   parallel  to   the  conjugate  or   revolving 
axis  ?  Ms.  517.1306. 


id  18; 

#4*  X  6 - 

oTvins 


190  Mensuration  of  Solids. 

To  find  the  Solidity  of  the  middle  Zone  of  a  Spheroid, 
having  circular  ends  ;  the  middle  Diameter,  and  that 
of  either  of  the  eitds  being  given. 

RULE, 

To  twice  the  square  of  the  middle  diameter,  add  the 
square  of  the  diameter  of  the  end;  multiply  the  sum 
by  the  length  of  the  zone,  and  the  product  again  by 
.2618  for  the  solidity. 

Note.  This  rule  is  useful  in  guaging.  A  cask  in 
the  form  of  a  middle  zone  of  a  prolate  spheroid,  is  by 
guagers  called  a  cask  of  the  first  variety. 

EXAMPLES. 

1.  What  is  the  solidity  of  the  middle  zone  of  a  pro- 
late spheroid,  the  diameter  am  of  the  end  being  36, 
the  middle  diameter  Nw  60,  and  the  length  B6  80  ? 

(2N?i2-j-am2)xB&X.26ls=  177940.224,  the  solid 
content. 

2.  Required  the  solidity  of  the  middle  zone  of  an 
oblate  spheroid,  the  middle  diameter  being  100,  the 
diameter  of  the  end  80,  and  the  length  36  ? 

Ans.  248814.72. 

3.  Required  the  content  in  ale  gallons  of  spheroidal 
cask,  whose  length  is  40  inches ;  the  bung  diameter 
being  32,  and  head  diameter  24  inches.  A  gallon  of 
ale  being  282  cubic  inches  ?  Jlns.  97.44. 

4.  Required  the  content  in  wine  gallons  of  a  spher- 
oidal cask,  whose  length  is  20  inches  ;  the  bung  diam- 
eter being  16,  and  head  diameter  12  inches.     A  gallon 


of  wine  being  231  cubic  inches  ? 


MS.  14.869, 


§  XIII  Of  a  Parabolic  Conoid. 

A  parabolic  conoid,  or  paraboloid,  is  a  solid,  gene- 
rated by  supposing  a  semiparabola  turned  about  its  axis. 


Mensuration  of  Solids. 
To  find  the  Solidity. 


191 


RULE. 


Multiply  the  square  of  the  diameter  of  its  base  by 
the  height,  and  that  product  by  .3927  ;  the  last  pro- 
duet  shall  be  the  solid  content. 


EXAMPLES. 


1.  Let  ABCD  be  a  par- 
abolic conoid,  the  diameter 
of  its  base,  AB,  36  inches, 
and  its  height,  CD,  33 
inches  ;  the  solidity  is  re- 
quired ? 


AB2X  CD  X. 3927=  16794.9936,  the  solid  content 
which  divided  by  172S  gives  9.719  feet  the  solidity. 

Demonstration  of  the  rule.  The  parabolic  co- 
noid is  constituted  of  an  infinite  number  of  circles, 
whose  diameters  are  the  ordinates  of  the  parabola. 
Now  according  to  the  property  of  every  parabola,  it 

will  be,  

AB» 

SA  :  AB  : :  AB" : =L,  the  Lotus  Rectum, 

SA 

rSaxL==ba* 
Then-]  SeXL=/f?a 

ISyxL-^S  $c. 
Here    SftxL,    HexL,   »%XL, 
&c.  are  a  series  of  terms  in  arith- 
metical progression.    Therefore 

ba2,  /e2,  gyz,  &c.  are  also,  a 
series  of  terms  in  the  same  pro- 
gression, beginning  at  the  point 


B,  wherein  AB2  is    the 


greatest 


.:•<£ 


i9~  JWetituration  uf  Solid*. 

term,  and  S A.  the  number  of  all  the  terms*.  There- 
fore, AB1  Xl*A=rthe  sum  of  all  the  terms  :  and  cir- 
cles are  to  each  other  as  the  squares  of  their  radii. 

Therefore,  if  we  put  D=AB,  and  H=8A. 

Thru  .78M  DDXiH=.3927  1)1)11  will  he  the  so- 
lid content  of  the  conoid:  which  is  just  half  the  cy- 
linder, whose  has?  i&=D,and  heights  H. 

This  being  rightly  understood,  it  will  he  easy  to 
raise  a  theorem  for  finding  the  lower  frustum  of  any 
parabolic  conoid. 

*  For  supposing  /i=«A,  the  height  of  the  frustum, 
and  o=S«,  the  height  of  the  part  b*b  cut  off,  then 
h+p=SA,  the  height  of  the  whole  conoid;  lctD  =  BB 
and  d=bb,  then  the  solidity  of  BSB  will  be  D2/t+ 
l)2;;;  X-3927,  and  the  solidity  of  bSb  will  he  d^px 
.3927,  hence  the  solidity  of  BMB 
must  be  D2/i-|-l)2/; — d2p  ;X-3'.>27,  & 

hut,  by  the  property  of  the  para-  /i    N« 


hola   SA  :  Sa  :  :  AB*  :  ab2,  viz.       I 


1)2  cF  hd* 


7 

i 


t>tl 


h+p:  p::  — :  —  hence  p  = /      v  \ 

4     4  1)*— d*      /     \ 

for p substitute  its  value  in  the  ex-  qjk" 

pression    for  the   solidity  of    the      v*- " 

hd2 

frustum  BMB,  and  it  becomes  (I)2/*;-f-Da — d*X ) 

D2— d* 

X  .3927=D2-f  r^2-f /iX.3927. 

2.  What  is  the  solidity  of  a  parabolic  conoid  whose 
height  is  84,  and  the  diameter  of  its  base  48  ? 

Jns.  76001.5872. 

3.  Required    the    solidity    of  a  paraboloid    whose 
height  is  30,  and  the  diameter  of  its  base  40  ? 

Ms.  18S49.6. 

4.  Required    the  solidity    of   a   paraboloid   whose 
height  is  22.S5,  and  the  diameter  of  its  base  32  ? 

Ms.  9188.55168. 


Mensuration  of  Solids.  193 


To  find  the  Solidity  of  the   Frustum  of  a   Paraboloid, 
or  Parabolic  Conoid. 

RULE. 

Multiply  the  sum  of  the  squares  of  the  diameters  of 
the  two  ends  by  the  height,  and  that  product  by  .3927 
for  the  solidity. 


Note.  This  rule  is  useful  in 
cask-gauging.  For  if  two  equal 
frustums  of  a  parabolic  conoid 
B&6B,  be  joined  together  at  their 
greater  bases,  they  form  a  cask  of     ^p.  JBPf 

the  third  variety.     The  less  dia-  B  ~ 

meter  of  the  frustum  bb  will  be  the  head  diameter  of 
the  cask,  and  the  greater  diameter  BAB  will  represent 
the  bung  diameter. 

EXAMPLES. 


1.  The  greater  diameter  BAB,  of  the  frustum  B6&B 
of  a  paraboloid  is  32,  the  less  diameter  bb  26,  and  the 
height  Aa,  8;  required  the  solidity  ? 

(BB2  4-^2)  xAax  .3927—5340.72,  the  solidity. 

2.  Required  the  solidify  of  the  frustum  of  a  para- 
bolic conoid,  whose  greater  diameter  is  30,  less  dia- 
meter 24-,  and  the  height  9  ?  Jins.  5216. 626S. 

3.  Required  the  solidity  of  the  frustum  of  a  para- 
boloid, the  diameter  of  the  greater  end  beins:  60,  that 
of  the  less  end  4*3,  and  the  height  13  ? 

Jlns.  41733.0144. 

4.  There  is  a  cask  in  the  form  of  two  equal  frustums 
of  a  parabolic  conoid ;  the  length  is  40  inches,  the 
bung  diameter  32,  and  head  diameter  24.  Required 
its  content  in  ale  gallons  ?  282  cubic  inches  being  one 
gallon.  Jlns.  89.1234  gallons. 

5.  There  is  a  cask  in  the  form  of  two  equal  frus- 
tums of  a  paraboloid :  the  length  is  20  inches,  the  bung 

S 


194 


Mensuration  of  Sol'nls. 


diameter  Ifi,  and  head  diameter  1  >.  Required  the  eon 
tent  in  wine  gallons  i  zm  cubic  inches  being  one  gal 
Ion.  Ans.  u.fi  gallons. 


§  XIV.     Of  olVarabolic  Spinels. 

A  parabolic  spindle  is   formed  by  the  revolution  of 
parabola  about  its  base,  or  greatest  double  ordinate. 


To  find  the  Solidify. 


RULE. 


Multiply  the  square  of  the  middle  diameter  by 
.11888  (being  jT  of  .7854)  and  that  product  by  its 
length;  the  last  product  is  the  solid  content 


EXAMPLES. 


1.  Let  ABCI)  be  a  parabolic  spin- 
dle, whos^  middle  diameter  CD  is  36 
inches,  and  its  length  AB  99  inches; 
fhe  solidity  is  required  ?  ^ 

CDaXABx.4-lSS=53743.979;52, 
inches,  or  31.10184  feet,  the  solidity. 


Mensuration  of  Solids.  195 

2.  The  length  of  a  parabolic  spindle  is  60,  and  the 
middle  diameter  34 ;  what  is  the  solidity  ? 

JhlS.  29053.516S. 

3.  The  length  of  a  parabolic  spindle  is  9  feet,  and 
the  middle  diameter  3  feet;  what  is  the  solidity  ? 

Jim.  33.92928. 

4.  What  is  the  solidity  of  a  parabolic  spindle,  whose 
length  is  40,  and  middle  diameter  16  ? 

Ms.  4289.3312. 

Demonstration  of  the  rule.  A  parabolic  spin- 
dle is  constituted  of  an  infinite  number  of  circles, 
whose  diameters  are  all  parallel  to  the  axis  SA  of  the 
parabola,  as  m  a,  n  e,py,  $*c. 

Let  us  suppose  the  line  Sa  parallel  to  AB,  &c. — 
Then  it  has  already  been  proved,  (section  XIV.)  that 
the  lines/  m, 
gii,  h  p,  Sfc. 
are  a  series  of 
squares,  whose 
roots  are  in  a- 
rithmetical  pro-  r  ^ 
gression,  consequently  their  squares,  viz.fm*,gn2,hp'ii 
ifr&.  will  be  the  series  of  biquadrates,  whose  roots  will 
be  in  arithmetical  progression :  which  being  premised, 
we  may  proceed  thus  : 

1.  &A—fm=.mu 

2.  SA — gn—ne 

3.  SA — hpzzpyi  4"c-  and  squaring  each  equation,  we 
set 


1.  &Ai--2$AXfm+fniz=ma2 

2,  SA2 — 2HAXgn+gn*=nei 


3.  S\*—2SAxhp+hp*=py?  $c. 

The  sum  of  these  equations  will  evidently  give  the 
sum  of  the  squares  of  the  radii,  m  a,ne,n  y,  <jfc.  of 
eircles  which  constitute  the  solidity  of  the  semi-nara- 
btlic  spindle.  The  first  terms  of  the  equations  being  a 


10G  Mensuration  of  Solids. 

-  <>f  equal  square*,  where  the  number  of  terms  is 
AB,  the  sum  of  the  terms  will  be  SpXAB. 

The  sum   of  the    second   terms   of  th<  qua- 

(jftfit  {Emu-son's  arithmetic  of  infinites,   PropJ  II!.)  is 
8A4.AB  L*xAB 

— 2SAx = wherein  SA  is  the 

3  8 

greatest  term,  and  AB  the  number  of  terms. 

The  sum  of  the  third  terms  of  the  above  equations 
(Emerson's  arithmetic  of  infinites,  Prop.  V.)  is 

l SA'XAB 

XSVxAB= jforfni'',  gn*,  hp*9  $C. 

4-j-l  5 

is  a  scries  of  biquadrates,  whose  greatest  term  is 

and  number  of  terms  AB.     Hence,  the  sum  of  all  the 

terms  of  .the  above  equations  will  be 

2SA*xAB    SA«XAB      88  A «      

SA2  x  AB ;  — + = X  AB, 

8  5  15 

the  sum  of  all  the  series  of  squares,  nut2,  ne*9pjf*i  #c- 
But  as  circles  are  to  each  other  as  the  squares  of  their 
radii  or  diameters  ;  it  evidently  follows  that  ,8T  SA* 
X-7SJ4XAB  will  he  the  solidity  of  one-fourth  of  the 
spindle  ASB,  that  is  T8T  of  the  cylinder  SaBA  circum- 
scribing one-fourth  of  the  spindle;  therefore  the  whole 
spindle  ty  °*  lis  circumscribing  cylinder  :  hence  if  D 
rzithe  middle  jdiameter,  and  L  the  length  of  the  spin- 
dle, its  solidity  will  be  ^Xl)2  XLX^834zzDa  X 
.41888X1'.     Which  is  the  rule. 


To  find  the  Solidity  of  one-fourth  of  the  middle  Fnis* 
turn  of  the  Spindle,  SpyA. 

Exactly   on    the    same   principles    as    above    SA2 

2$Axhp                W^ 
X  Ay  ;- — |-  Ay  ;-j .  X  Ay=(sum  of  all  the 


Mensuration  of  Solids.  19^ 

series     of   squares    8Aa,    ma'1,   -we5,    and  ^#%)= 

2SAX/?,w      tip*  \ 

SA* 1+ )  

3  5    /XAt/=S.    Hence  3SA8—- 

3hp     3S 


2SAX/^H = — •      But   SAa—  2SAx/^+/ip2~- 

g        Ay 
^y*,  and   subtracting  this   equation  from  the  former 

ify?         3S      

we  get  2$ A* -J %2=z i>#%  hence  by  re- 
duction, &c.         5                      Ay 

2SA*-f;»y2— f/*/>2 ;  XjApS.     And    circles   are   to 
each  other  as  the  squares  of  their  radii,  hence 

1.5708SA24-.7S54py2   —  3.i4>16hpz  X  i  Ay=S,     or 

2SA4-fjoy2 — lfyo2X.261S  Ay=8,  the  solidity  of  one- 
fourth  of  the  middle  frustum  of  the  spindle.   Hence 

To, find  the  Solidity  of  the  middle  Frustum  of  a  Para- 
bolic Spindle. 

To  twice  the  square  of  the  middle  diameter,  add 
the  square  of  the  diameter  of  the  end;  and  from  the 
sum  subtract  four-tenths  of  the  square  of  the  differ- 
ence between  these  diameters;  the  remainder  multi- 
plied by  the  length,  and  that  product  by  .2618  will 
give  the  solidity. 

Note.  This  rule  is  useful  in  cask-gauging.  A  cask 
in  the  form  of  the  middle  frustum  of  a  parabolic  spin- 
dle, is  called  by  gaugers  a  cask  of  the  second  variety ; 
and  is  the  most  common  of  any  of  the  varieties. 

EXAMPLES. 

1.  Required  the  solidity  of  the 
middle  frustum  of  a  parabolic 
spindle  EFGH,  the  length  AB 
being  20,  the  greatest  diameter 
CD  16,  and  the  least  diameter 
EForGHl2? 

s  2 


Mensuration  of  Sol 

D  -f-GH*  —  .4X(CI)  — GH)2)xAlix.-'« 
(5134-144  —  .4X4X4) >C«0  X  .S618  =  ($iW  —  6.4)  X 
.■>.2i6  =  640.6X3.236 =3401.3056,  the  solid  cont; 

2.  The  bung  diameter  CD  of  a  cask  is  32  inches, 
head  diameter  EF,  21  inches,  and  length  AB,  40 
inches  ;  required  its  content  in  ale  gallons  ?  282  cubic 
inches  being  t  gallon.  Ans,  96.4909  gallons. 

3.  Tl:e  bung  diameter  (  1),  of  a  cask  is  JO  inches, 
head  diameter  EF  20  inches,  and  length  AB  36  inches; 
required  its  content  in  nine  gallons?  231  cubic  inches 
being  1  gallon.  Ans.  117.89568  gallons. 

To  find  the  Solidify  of  the  middle,  Frustum  of  any  Spin- 
die,  formed  by  the  revolution  of  a  Conic  Section  about 
the  Diameter  of  that  Section. 


To  the  square  of  the  greatest  diameter  add  the 
square  of  the  least,  and  four  times  the  square  of  a 
rliameter  taken  exactly  in  the  middle  between  the  two'; 
multiply  the  sum  by  the  length,  and  that  product  by 
,1309  for  the  solidity. 

Note.  See  the  latter  part  of  the  demonstration  of 
the  rule,  iu  section  X. 

EXAMPLES. 

1 .  Required  the  solidity  of  the  middle  frustum  EFGH 
of  any  spindle ;  the  length  All,  being  40,  the  greatest 
or  middle  diameter  CD,  32,  the  least  diameter  EF  or 
Gil,  24,  and  the  diameter  IK  in  the  middle  between 
GH  and  CD,  mt37568  ?  Ans.  2742.3.72624. 

2.  The  bung  diameter  of  a  cask  being  36  inches  ; 
head  20;  length  36,  and  a  diameter  exactly  in  the 
middle,  31.93  inches;  what  is  the  content  in  wine 
gallons  ?  Ans.  117  gallons  3*  quarts'. 


Mensuration  of  Regular  Bodies.  190 


§  XV.  Of  the  five  Regular  Bodies. 

A  regular,  or  pi  atonic  body,  is  a  solid  contained 
under  a  certain  number  of  similar  and  equal  plane 
figures.  Only  three  sorts  of  regular  plane  figures 
joined  together  can  make  a  solid  angle ;  for  three  plane 
angles,  at  least,  are  required  to  make  a  solid  angle, 
and  all  the  plane  angles  which  constitute  the  solid 
angle,  must  be  less,  when  added  together,  than  four 
right  angles,  {Euclid,  XL  and  21.)  Now  each  angle  of 
an  equilateral  triangle  is  60  degrees ;  each  angle  of  a 
square  90  degrees;  and  each  angle  of  a  pentagon  180 
degrees.  Therefore  there  can  be  only  five  regular 
bodies,  for  the  solid  angles  of  each  must  consist  either 
of  three,  four,  or  jive  triangles,  three  squares,  or  three 
pentagons. 

1.  The  tetraedron,  or  equilateral  pyramid,  which  has 
four  triangular  faces ;  hence  all  the  plane  angles  about 
one  of  its  solid  angles  make  180  degrees. 

2.  The  octaedron,  which  has  eight  equilateral  trian- 
gular faces;  hence  all  the  plane  angles  about  any  one 
ol  its  solid  angles  make  240  degrees. 

3.  The  dodecaedron,  which  has  twelve  equilateral 
pentagonal  faces ;  hence  all  the  angles  about  any  one 
of  its  solid  angles,  make  324  degrees. 

4.  The  icosaedron,  which  has  twenty  equilateral  tri- 
angular faces ;  hence  all  the  angles  about  any  one  of 
its  solid  angles,  make  300  degrees. 

5.  The  he.vacdron,  or  cube,  which  has  six  equal 
square  faces ;  hence  all  the  angles  about  any  one  of  its 
solid  angles  make  270  degrees. 

Note.  If  the  following  figures  be  exactly  drawn  on 
pasteboard,  and  the  lines  cut  half  through,  so  that  the 
parts  may  be  turned  up  and  glued  together,  they  will 
represent  the  five  regular  bodies  above  described. 


.200 


The  Mensuration  df 


or  i  /* 


Ifie-lcojaedron, 


the  Five  Regular  Bodies. 


201 


A  table,  shewing  the  solidity  and  superficies'  of  the 
five  regular  bodies,  the  length  of  a  side  in  each  be- 
ing 1,  or  unity. 


JSTames  of  the  bodies. 

Solidity. 

Superficies. 

Tetraedron 

.11785113 

205081 

Oetaedron 

.47110-152 

8.46410162 

Hexaedron 

1.00000000 

6.00000000 

Icosaedron 

2.18 169499 

8.6602540-1 

Dodeeaedron 

7.6631189 

20.6457288 

To  find  the  Solidity  or  Superficies,  of  any  regular 
Body,  by  the  Table. 


RULE, 


1.  Multiply  the  cube  of  the  length  of  a  side  of  the 
body,  by  the  tabular  solidity,  and  the  product  will  give 
the  solidity  of  the  body. 

2.  Multiply  the  square  of  the  length  of  a  side  of 
the  body,  by  the  tabular  superficies,  and  the  product 
will  give  the  siperficies  of  the  body. 


EXAMPLES. 


1.  Let   ABCD   be    a  tetraedron,  /  j  \ 

whose  side  AB  is  12  inches;  requir-  /   (    \ 

ed  the  solidity  and  superficies  ?  /     g  \ 


12=AB 

12 


Tab.  sol.     .11785113 

1728 


144= square  of  AB 

12 

1728= cube  of  AB 


94280904 
23570220 
82195791 
11785113 

203.64675264  solidity. 


The  Mensuration  of 

And.  ill-  multiplied  by  t. 78205  081,  the  tabular  su 
perlicies,  gives  249.41531664  inches,  the  superfieio  oj 
the  tetraedron. 

2.  Each  side  of  a  tetraedron  is  3,  required  its  sur 
face  and  solidity  ? 

Atis,  Superficies  15.58843729;  solid.  3.18198051 

a.LetABCDE  be  an  ottae- 

dron,  each  side  being  L2  inch- 
es; required  the  solidity  and 
superficies  ?  *  / 

Answer. 

814.58701056  inch,  solid. 

49S.S306332S  inch,  super. 


4.  Let  B  be  a  h&vaedron,  or  cube, 
whose  side  is  12 inches;  required  the 
solidity  and  superficies? 

a        C1728  inches,  solidity. 

'  \    864  inches,  superiieies.       [ 


5.  Let  ABCDEPGH1     fr's 

he  an  icosaedron,  each  side 
thereof  being  12  inches  ;  j    \ 

required  the  solidity  and 
superficies  ? 

■  Hj 

& 

a        5  3769.97470272  inches,  solidity. 
*  1 1247.07638176  inches,  superficies. 


VI  \ 

\ 

V     -V 

the  Five  Regular  Bodies. 


203 


6.  Let  ABCDEFGHIK  be  a  dodecaedron,  each  side 
thereof  being  12  inches;  required  the  solidity  and 
superficies  ? 

a        C  13241.8694592  inches,  solidity. 

1  '  £   2972.9849472  inches,  superficies. 


§  XVI.  To  measure  any  Irhegulam  Solid. 

If  you  have  any  piece  of  wood  or  stone  that  is  craggy 
p  uneven,  and  you  desire  to  find  the  solidity,  put  the 
solid  into  any  regular  vessel,  as  a  tub,  a  cistern,  or 
;he  like,  and  pour  in  as  much  water  as  will  just,  cover 
t;  then  take  out  the  solid,  and  measure  how  much  the 
pall  of  tho  water  is,  and  so  find  the  solidity  of  that 
part  of  the  vessel. 

EXAMPLES. 

1.  Suppose  a  piece  of  wood  or  stone  to  he  measured, 
md  suppose  a  cylindrical  tub  32  inches  diameter,  into 
which  let  the  stone  or  wood  be  put,  and  covered  with 
.vater:  then,  when  the  solid  is  taken  out,  suppose  the 
all  of  the  water  14  inches  ;  square  32,  and  multiply 
the  square  by  .7854,  the  product  will  be  804.2496,  the 


:oi  The  Mensuration  of  §c. 

area  of  the  base  ;  which  multiplied  by  14,  the  depth 
or  fall  of  the  water,  the  product  is  11359.40,  &c.  which 
divided  by  t:  38,  the  quotient  is  0.31  feet;  and  so  much 
is  tin*  solid  content  required. 

2.  It  is  reported  that  Hiero,  king  of  Sicily,  furnish- 
ed a  workman  with  a  quantity  of  gold  to  make  a 
crown.  When  it  came  home,  lie  suspected  that  the 
workman  Itad  used  a  greater  alloy  of  silver  than  was 
necessary  in  its  composition,  and  applied  to  Archi- 
medes* a  celebrated  mathematician  of  Syracuse,  to  dis- 
cover the  fraud,  without  defacing  the  crown.  He  pro- 
cured a  mass  of  gold  and  another  of  silver,  exactly  of 
the  same  weight  with  the  crown ;  considering  that  if 
the  crown  were  of  pure  gold,  it  would  be  of  equal 
hulk  and  displace  an  equal  quantity  of  water  with  the 
golden  ball:  and  if  of  silver,  it  would  be  of  equal 
bulk,  and  displace  an  equal  quantity  of  water  with  the 
silver  ball;  but  if  a  mixture  of  the  two,  it  would  dis- 
place an  intermediate  quantity  of  water. 

Now,  for  example,  suppose  that  each  of  the  three 
masses  weighed  loo  ounces;  and  that  on  immersing 
them  severally  in  water,  there  were  displaced  5  ounces 
of  water  by  the  golden,  9  ounces  by  the  silver,  and  G 
ounces  by  the  crown ;  then  their  comparative  bulks  are 
5,  9,  and  0. 

From  9,  silver.  From  6,  crown. 

Take  6,  crown.  Take  5,  gold. 

3  remainder.  1  remainder. 

The  sum  of  these  remainders  is  4; 
Then  4  :  100  ::  3  :  73  ounces  of  gold, 
and  4  :  100  ::  1  :  25  ounces  of  silver. 
That  is,  the  crown  consisted   of  75  ounces  of  gold; 
and  25  ounces  of  silver. 

See  Cotes'  Hydros.  Led.  p.  SI, 


The  Measuring  of  Boards  and  Planks.         205 

CHAPTER  III. 

The  Measuring  of  Board  and  Timber. 

§  I.     Of  Board  Measure. 
To  find  the  Superficial  Content  of  a  Board  or  Plank, 

RULE. 

MULTIPLY  the  length  by  the  mean  breadth.  When 
the  board  is  broader  at  one  end  than  the  other,  add 
the  breadths  of  the  two  ends  together  and  take  half 
the  sum  of  the  mean  breadth. 

By  the  Carpenter's  Rule. 

Set  12  on  B  to  the  breadth  in  inches  on  A ;  then 
against  the  length,  in  feet,  on  B,  you  will  find  the  su- 
perficies on  A,  in  feet. 

By  Scale  and  Compasses. 

Extend  the  compasses  from  12  to  the  length  in  feet, 
that  extent  will  reach  from  the  breadth,  in  inches,  to 
the  superficies  in  feet. 

EXAMPLES. 

1.  If  a  board  be  16  inches  broad,  and  13  feet  longj 
how  many  feet  are  contained  in  it  ? 

Multiply  16  by  13,  and  the  product  is  208 ;  which 
divided  by  12,  gives  17  feet,  and  4  remains,  which  is 
a  third  part  of  a  foot. 

T 


■.'06         The  M  of  Boards  and  1'lr . 

fly  the  Carj'Otfn's Jlitlc. 
As  12  on  B  :   16  on  A  :  :   13  on  B  :  17£  on  A. 

fly  Scale  and  Compasses. 

Extend  tlie  compasses  from  12  to  18,  the  length  in 
feet,  that  extent  will  reach  from  16,  the  hreadlh  in 
inches,  to  IT  \-  t  lie  superficies  in  feet. 

2.  What  is  the  value  of  a  plank,  whose  length  i->  s 
feet  0  inches,  and  breadth  throughout  1  foot  3  inches; 
at  2U\.  per  foot  ?  Ms.  2  shill.  2 hi 

3.  Required  the  superficies  of  a  board,  whose  mean 
breadth  is  1  foot  finches,  and  length  12  feet  6  inches  ? 

Jins.  14  feet  7  inches. 

4.  Having  occasion  to  measure  an  irregular  mahog- 
any plank  of  14  feet  in  length,  I  found  it  necessary  to 
measure  several  breadths  at  equal  distances  from  each 
other,  biz,  at  every  two  feet.  The  breadth  of  the  less 
end  was  6  inches,  and  that  of  the  greater  end  1  foot  ; 
the  intermediate  breadths  were  1  foot,  1  foot  6  inches, 
2  feet,  2  feet,  1  foot,  and  2  feet ;  how  many  square  feet 
were  contained  in  the  plank  ? 

Ans.  t&j  feet;  — meau  breadth  1|  feet,  found  by  divi- 
ding the  sum  of  the  breadths  by  their  number. 

5.  Required  the  value  of  5  oaken  planks  at  3d.  per 
foot,  each  of  them  being  17j  feet  long;  and  their  se- 
veral breadths  as  follows;  viz.  two  of  i8£  inches  in 
the  middle,  one  of  14V  inches  in  the  middle,  and  the 
two  remaining  ones,  each  IS  inches  at  the  broader  end, 
and  11  \  inches  at  the  narrower  ?         +1ns.  Li  :  5  :  H\. 


The  Measuring  of  Boards  and  Planks.       207 

Having  the  breadth  aj  a  rectangular  Plank  given  in 
inches,  to  find  how  much  in  length  will  make  afoot, 
or  any  other  assigned  quantity. 

Divide  144,  or  the  area  to  he  cut  off,  by  the  breadth 
in  inches,  and  the  quotient  will  be  the  length  in  inches. 

Note.  To  answer  the  purpose  of  the  above  rule, 
some  carpenters'  rules  have  a  little  table  upon  them, 
in  the  following  form,  called 

A  Table  of  Board  Measure. 


0 

0 

0 

0 

5 

0 

n 

6 

12 

6 

4 

3 

* 

2 

i 

1 

1 

2 

3 

4 

5 

6 

7 

8 

By  this  table  you  are  to  understand,  that  if  the 
breadth  be  1  inch,  the  length  must  be  12  feet ;  if  2 
inches,  the  length  is  6  feet ;  if  5  inches  broad,  the 
length  is  2  feet  3  inches,  &e. 

If  the  breadth  be  not  contained  in  the  table  on  the 
rule,  shut  the  rule,  and  look  for  the  breadth  in  the 
line  of  board  measure,  which  runs  along  the  rule  from 
the  table  of  board  measure,  and  over  against  it  on  the 
opposite  side,  in  the  scale  of  inches,  is  the  length  re- 
quired. Thus,  if  the  breadth  be  9  inches,  you  will 
find  the  length  against  15  inches  ;  if  the  breadth  be 
41  inches,  you  will  find  the  length  a  little  above  13 
inches,  &c. 

EXAMPLES. 

i.  If  a  board  be  19  inches  broad;  how  many  inch- 


in  length  will  make  a  foot  ? 


Am.  7.58  inches 


203  Tlie  Mensuration  of  Timber. 

2.  From  a  mahogany  plank  26  inches  broad,  a  yard 
and  a  half  is  to  be  cutott';  what  distance  from  the  end 
must  the  line  be  struck  ? 

.flws.  74.7692  inches,  or  6.23  icet 


§  II.  The  Customary  Method  of  Measuring 
Timber. 

I.  The  customary  method  of  measuring  round  tim- 
ber, is  to  gird  the  piece  round  the  middle  with  a  string  ; 
one  fourth  part  of  this  girt  squared  and  multiplied  by 
the  length  gives  the  solidity.  Or,  if  the  piece  of 
timber  be  squared,  half  the  sum  of  the  breadth  and 
depth  in  the  middle  is  considered  as  the  quarter-girt, 
and  used  as  above. 

II.  If  the  piece  of  timber  be  very  irregular,  gird  it 
in  several  places  equally  distant  from  each  other,  and 
divide  the  sum  of  their  circumferences  by  their  number, 
for  a  mean  circumference  5  the  square  of  one  quarter 
of  the  mean  circumference,  multiplied  by  the  length, 
will  give  the  solidity. 

Note.  If  the  circumference  be  taken  in  inches, 
and  the  length  it  feet,  divide  the  last  product  by  144. 


"The  Mensuration  of  Timber. 


209 


III.  Otherwise  by  the  following  Table  for  measuring 

Timber. 


Quart. 

Quart. 

Quart. 

girt. 
Inches. 

Area. 

girt. 

Area. 

girt. 

Area. 

Feet. 

Inches. 

Feet. 

Inches. 

Feet. 

6 

.230 

12 

1.000 

18 

2.250 

H 

.272 

i2\ 

1.042 

18  J- 

2.376 

6£ 

.294 

13$ 

i.085 

19 

2.506 

6| 

.317 

i$t 

1.129 

m 

2.640 

7 

.340 

13 

1.174 

20 

2.777 

a 

.364 

13* 

1.219 

202 

2.917 

*  i 

.390 

1*1 

1.265 

21 

3.062 

rt 

.417 

ia* 

1.313 

211 

3.209 

8 

.444 

14 

1.361 

22 

3.362 

8^ 

.472 

14  \ 

1.410 

2*1 

3.516 

»2 

.501 

141 

1.460 

23 

3.673 

s* 

.531 

l4| 

1.511 

2;>1 

3.835 

9 

.562 

15 

1.562 

24 

4.000 

K 

.594 

IS* 

1.615 

24' 

4.168 

9* 

.626 

±5\ 

1.668 

25 

4340 

ft 

.659 

15| 

1.722 

"°2 

4.5 1 6 

10 

.694 

16 

1.777 

26 

4.694 

10| 

.730 

16} 

1.833 

26j 

4.876 

10] 

.766 

iftj 

1.890 

27 

5.062 

iOf 

.803 

16| 

1.948 

sp  2 

5.252 

n 

.840 

17 

2.006 

28 

5.444 

Hi 

■     .878 

17  V 

2.066 

28  £ 

5.640 

Ui 

.918 

171 

2.126 

29 

5.840 

fit 

.939 

17| 

2.187 

29i 
30 

6.044 
6.230 

T   2 


tittt  The  Mensuration  of  Timber. 


Tlie  use  of  the  Table. 

Multiply  the  area  corresponding  to  the  quarter-girt  . 
in  inches,  by  the  length  of  the  piece  of  timber  in  feet. 
and  the  product  will  be  the  solidity. 

IV.  By  tlie  Carpenters  Rule. 

Measure  the  circumference  in  the  middle  of  the 
piece  of  timber,  and  take  a  quarter  of  it  in  inches,  call 
this  the  girt. 

Then  set  12  on  D  to  the  length  in  feet  on  C,  and 
against  the  girt  in  inches  on  D,  you  will  find  the  con- 
lent  in  feet  on  C 

V.  By  Scale  arid  Comjwsses. 

Measure  the  circumference  in  the  middle  of  the 
piece  of  timber,  and  take  a  quarter  of  it  in  inches. 

Then  extend  from  12  to  this  quarter-girt,  that  ex-' 
rent  will  reach  from  the  length  in  feet,  being  turned 
twice  over,  to  the  solidity. 

Note.  The  buyer  is  allowed  to  take  the  girt  any 
where  between  the  greater  end,  and  the  middle  of  the 
tree,  if  it  taper. 

All  branches,  or  boughs,  whose  quarter-girt  is  not 
less  than  six  inches,  are  reckoned  as  timber ;  and  any  \ 
part  of  the  trunk  less  than  2  feet  in  compass  is  not 
considered  as  timber. 

An  allowance  is  generally  made  to  the  buyer  on 
account  of  bark ;  thus  for  oak,  one-tenth  or  one-twelfth 
part  of  the  circumference  is  deducted;  but  the  allow- 
ance for  the  bark  of  ash,  beech,  elm,  &c.  is  a  small 
matter  less. 

EXAMPLES. 

1.  If  a  piece  of  square  timber  be  2  feet  9  inches 
deep,  and  1  foot  7  inches  broad  5  and  the  length  16 


■ 


The  Mensuration  of  Timber. 


211 


feet  9  indies,  (or,  which  is  the  same   thing,*  if  the 
I  quarter-girt  be  26  inches,  and  length  16  feet  9  inches) 
how  many  solid  feet  are  contained  therein  ? 


26  inches, 
26 

quarter 

ile. 

feet. 

-girt.             16.75=16  feet  9  inch 
676 

156 
52 

676  square. 

%  the  Ted 

4  694 

10050 
11725 
10050 

144)11323.00(7S.63  feet. 

1243 

16.75 

910 

23470 
32858 
28164 
4694* 

460 
28  remainder. 

78.62450 

Note.  The  true  content,  measured  as  a  parallel- 
opipedon,  §  II.  of  chap.  II.  is  72.93  feet.  See  example 
4,  page  137. 


*  That  is,  according-  to  the  customary  method  of  measuring", 
mentioned  at  the  beginning  of  this  section.  And  when  the 
breadth  and  depth  of  a  piece  of  timber  are  nearly  equal,  this 
method  is  nearly  true. — Otherwise,  the  breadth  in  the  middle,  is 
generally  multiplied  by  the  depth  in  the  middle,  and  that  product 
by  the  length. 

When  the  timber  tapers  regularly,  half  the  sum  of  the 
breadths  of  the  two  ends  is  the  breadth  in  the  middle,  and 
half  the  sum  of  the  depths  of  the  two  ends  is  the  depth  in 
the  middle. 


:i-  The  Mensuration  of  Tin 

liy  the  Carpenter's  Rule. 

As  12  on  D  :  16£  on  C  : :  26  on  D  :  78£  an  C. 

/?</  &ca/<'  and  Comjmsses, 

Extend  from  12  to  26,  that  extent,  being  turned 
twice  over,  will  reach  from  16f  feet  to  78  feet. 

2.  The  quarter-girt  of  a  piece  of  squared  timber  is 
13  inches,  and  the  length  18  feet,  required  the  so- 
lidity ?  Jlns.  28£  feet. 

3.  If  a  piece  of  squared  timber  be  25  inches  square 
at  the  greater  end,  and  9  inches  square  at  the  less, 
(or,  which  is  the  same  thing,  the  quarter-girt  in  the 
middle  be  17  inches)  and  the  length  20  feet ;  how 
many  feet  of  timber  are  contained  therein  ? 

Ans.  40.13  feet/ 
Note.  The  true  content,  measured  as  the  frustum 
of  a  square  pyramid,  §  VII.  chap.  II.  is  43.101  feet. 
See  example  4,  page  162. 

4.  If  a  piece  of  squared  timber  be  32  inches  broad 
and  20  inches  deep  at  the  greater  end,  and  10  inches 


The  quarter-girt  rule  never  agrees  with  this,  except  the 
dimensions  of  a  piece  of  timber  in  the  middle  be  a  true  square. 
In  every  other  case,  when  the  timber  is  tapering,  this  rule 
gives  the  content  too  little  .-  and  the  quarter-girt  rule  is  nearer 
the  truth,  until  six  times  the  square  of  the  quarter-girt,  be 
exactly  equal  to  the  sum  of  the  areas  of  .the  two  ends,  toge- 
ther with  an  area  found  by  multiplying  the  sum  of  the  breadths 
of  the  ends  by  the  sum  of  their  depths ;  and  then  it  is  exactly 
true.  Thus,  in  the  8th  example  following,  the  quarter-girt 
rule  is  nearer  the  truth  than  the  rule  given  in  tins  note  :  but 
when  a  piece  of  timber  tapers  very  little,  and  its  breadth  dif- 
fers materially  from  the  depth,  the  quarter-girt  rule  ought  to 
be  rejected,  and  this  rule  in  the  note  should  be  used  instead 
thereof. 


The  Mensuration  of  Timber.  213 

broad  and  6  inches  deep  at  the  less  end,  (or,  which  is 
the  same  thing,  the  quarter-girt  in  the  middle  be 
17  inches)  and  the  length  18  feet;  how  many  feet  of 
timber  are  contained  therein  ?  Jlns.  36.11,  &c.  feet. 
Note.  The  true  content  by  §  VII.  chap.  II.  rule  1, 
is  37.33  feet.     See  example  5,  page  162. 

5.  If  a  piece  of  round  timber  be  96  inches  in  cir- 
cumference, or  the  quarter-girt  be  24  inches,  and  the 
length  18  feet;  how  many  feet  of  timber  are  contain- 
ed therein  ?  Jlns.  72  feet. 

Note.  The  true  content,  measured  as  a  cylinder  by 
§  V.  chap.  II.  is  91.67  feet.    See  example  2,  page  149. 

6.  If  a  piece  of  round  timber  be  86  inches  in  cir- 
cumference, or  the  quarter-girt  be  21  £  inches,  and  the 
length  20  feet ;  how  many  feet  of  timber  are  contained 
therein  ?  Jlns.  64.2  feet. 

Note.  The  true  content,  measured  as  a  cylinder  is 
81.74  feet.  See  example  3,  page  149. 

7.  If  a  piece  of  round  timber  be  28.2744  inches  in 
circumference  at  the  less  end,  and  113.0976  inches  in 
circumference  at  the  greater  end,  (or,  which  is  the 
same  thing,  the  quarter-girt  in  the  middle  be  17.6715 
inches)  "and  the  length  24  feet ;  how  many  feet  of 
timber  are  contained  therein  ?  Jlns.  52.047  feet 

Note.  The  true  content,  measured  as  the  frustum  of 
a  cone,  by  §  VIII.  chap.  II.  is  74.22  feet.   Vide  p.  163. 

8.  If  a  piece  of  timber  be  136  inches  in  circumfer- 
ence at  the  greater  end,  and  32  inches  in  circumference 
at  the  less  end,  (or,  which  is  the  same  thing,  the  quar- 
ter-girt in  the  middle  be  21  inches)  and  the  length  21 
feet ;  how  many  feet  of  timber  are  contained  therein  ? 

Jlns.  64.31  feet. 
Note.  The  true  content,  measured  as  the  frustum 
of  a  cone,   by   §  VIII.  chap.   II.    92.34  feet.       Vide 
page  163. 

9.  How  many  solid  feet  are  contained  in  a  tree 
whose  length  is  17^  feet,  and  girts  in  five  places,  viz. 
in  the  first  place  9.43  feet,  in  the  second  7.92,  in 


,:i*  TV  Mensuration  of  Timber. 

in  the  first   place    <>.4J    feet,    in  the   second  T.92,  in 
the  third,  tf.i59  in  the  fourth  4.74,  and  in  the  fifth 
Feet  ?  Jns.  42.52  feet. 

To  find  how  much  in  length   ivill  make  a  foot  of  any 
squared  Timber,  of  equal  thickness  from  end  to  end. 

Divide  1728,  the  solid  inches  in  a  foot,  by  the  area 
of  the  end  in  inches,  and  the  quotient  will  be  the 
length  of  a  solid  foot,  in  inches. 

Note.  To  answer  the  purpose  of  the  above  rule, 
the  carpenters'  rules  sometimes  have  a  little  table  up- 
on them  in  the  following  form,  called 

A  Table  of  Timber  Measure. 


O     |     0| 

o 

» 

■■• 

[•; 

11  |  3 

|  9  j  Inches. 

14*  |  36 

15 

• 

» 

* 

2\2 

1  |  Feet. 

*     1     ».l 

»| 

* 

? 

•  1 

r|sj 

9  j  hide  of  the  sq. 

By  this  table  you  are  to  understand,  that  if  the  side 
of  the  square  be  one  inch, the  length  must  be  14  f  feet; 
if  2  inches  be  the  side  of  the  square,  the  length  must 
be  36  feet,  to  make  a  solid  foot. 

If  the  side  of  the  square  be  not  in  the  little  table, 
look  for  it  in  the  line  of  timber  measure,  running  along 
the  rule  from  the  table,  and  against  it,  in  the  line  of 
inches  is  the  length  required.  Thus,  if  the  side  of  the 
square  be  16  inches,  you  will  find  the  length  to  be  6 
inches  and  7  tenths,  &c. 

EXAMPLES. 

1.  If  a  piece  of  timber  be  18  inches  square,  how 
much  in  length  will  make  a  solid  foot  ? 

.fyis.  ."U  inches. 


The  Mensuration  of  Timber,  215 

2.  Ti"  apiece  of  timber  be  22  inches  deep,  and  15 
Riches  broad  :  how  much  in  length  will  make  a  solid 
toot  ?  Sins.  5.23  inches. 


SCHOLIUM. 

The  foregoing  section  contains  the  whole  substance 
of  timber  measuring,  as  practised  in  all  timber  yards ; 
and  though  the  method  has  been  noticed  as  erroneous 
by  various  different  writers  for  near  a  century  past,  it 
still  stands  its  ground  ;  nor  does  it  appear  probable 
that  it  will  soon  be  abolished. 

In  measuring  round  timber  by  the  foregoing  method, 
of  taking  a  quarter  of  the  circumference  in  the  mid- 
dle, for  the  side  of  a  mean  square ;  it  is  objected  that 
it  makes  the  content  too  little,  and  that  such  timber 
ought  to  be  measured  as  a  cylinder,  or  the  false  con- 
tent should  be  increased  in  the  ratio  of  .0625  to  .0795S, 
or  as  11  to  14. 

But  this  objection  is  answered  by  saying,  that  be- 
fore the  wood  can  be  squared,  and  made  fit  for  use,  a 
great  part  of  it  goes  to  waste  ia  chips,  and  therefore 
the  quantity  of  round  timber  ought  to  be  reckoned  no 
more  than  what  the  inscribed  square  will  amount  to. 
Xow,  if  the  circumference  be  1,  the  area  of  the  sec- 
tion will  be  .0795S,  the  square  of  the  quarter-girt 
.0625,  and  the  side  of  the  inscribed  square  .22">1,  the 
square  of  which  is  .05067.  Therefore  to  make  the 
content  by  the  inscribed  square  correspond  with  the 
cylindrical  content,  it  ought  to  be  increased  in  the  ra- 
tio of  .03067  to  .07958,  or  7  to  11  which  is  a  greater, 
increase  than  1 1  to  11,  and  therefore  instead  of  the 
content  by  the  quarter-girt  being  too  little,  it  is  on  this 
consideration  too  much. 

Again,  when  you  take  the  mean  girt  of  a  tree,  it  is 
very  probable  that  the  tree  is  not  perfectly  circular 
in  that  place,  and  the  more  it  differs  from  a  circle 
the  greater  the  quarter-girt  will  be ;  so  that  if  you 


216  Tlie  Mensuration  of  Timber. 

were  to  measure  such  a  tree  as  a  cylinder,  by  the  pro- 
per rule  for  that  purpose,  the  content  Mould  in  reality 
be  too  much  on  account  of  the  error  in  taking  the 
girt. 

It  is  likewise  objected  that  in  round  tapering  tim- 
ber, taking  the  side  of  a  square  in  the  middle  of  the 
piece  makes  the  content  too  little,  and  that  even  in  a 
greater  proportion  than  above.  The  answer  given 
to  this  objection  is,  that  in  almost  in  all  cases  the  great-  ' 
er  must  be  cut  away  till  it  be  of  the  same  dimensions 
as  the  less  end,  otherwise  the  timber  cannot  be  sawn 
into  useful  materials  ;  and  therefore  there  is  no  just 
reason  for  any  objection  to  the  rule  on  this  ground ; 
and  particularly,  as  no  general  rule  has  hitherto  been« 
published,  of  sufficient  merit,  to  supersede  the  use  of 
the  quarter-girt  rule. 

Dr.  Ifutton,  in  the  quarto  edition  of  his  Mensura- 
tion (published  in  17/0)  p.  607,  has  given  an  easy 
rule  for  measuring  round  timber  :  thus, 

"  Multiply  the  square  of  one-fifth  of  the  girt  or  ctr- 
cumference  by  twice  the  length,  and  the  product  will  be 
the  content  {extremely  near  the  truth.)" 

Dr.  Hutton  says  that  many  reasons  may  be  alledg- 
ed  for  changing  the  customary  method  of  measuring 
timber,  and  introducing  this  rule  instead  thereof,  and 
the  principal  reason  is,  (see  p.  6 14  of  the  quarto  edi- 
tion of  his  Mensuration)  "  the  preventing  of  the  sel- 
lers from  playing  any  tricks  with  their  timber  by  cut- 
ting trees  into  different  lengths,  so  as  to  make  them 
measure  to  more  than  the  whole  did  ;  for,  by  the  false 
method,  this  may  be  done  in  many  respects."  Mr. 
Mpnnycastk  gives  the  same  problems,  "  to  shew  the 
artifices  that  may  be  used  in  measuring  timber  ac- 
cording to  the  false  method  now  practised,  and  the 
absolute  necessity  there  is  for  abolishing  it." 

It  is  rather  singular  that  neither  of  those  gentle- 
men should  perceive,  that  the  very  same  tricks  and 
artifices  may  be  practised,  with  equal  success,  if  Dr. 


The  Mensuration  of  Timber.  217 

Hutiotfs  rule  be  used.  The  truth  of  these  remarks 
will  easily  appear  to  those  who  are  qualified  to  read 
the  Doctor's  demonstrations;  and  those  who  are  not, 
may  consult  the  following 

EXAMPLES. 

1.  Dr.  Hutton,  p.  613,  Prob.  IV. 
Supposing  a  tree  to  girt  14  feet  at  the  greater  end, 
2  feet  at  the  less,  and  8  feet  in  the   middle;  and  that 
the  length  be  32  feet. 

Rale.  If  this  tree  be  cut  through,  exactly  in  the 
middle,  the  two  parts  will  measure  to  the  most  possi- 
ble, by  the  common  method,  and  to  more  than  the 
whole. 

By  the  common  Method. 

The  whole  tree  measures  1.28  feet. 

The  sreater  end  measures  121  feet. 

The  tess  end  measures  25  feet. 


ing  the  whole  by  18  feet. 


Sum  146    exceed 


By  Dr.  Hutton' 's  Rule. 
The  whole  tree  measures  163.84  feet. 


The  greater  end  measures  154.88  feet. 

The  less  end  measures  32.       feet. 


Sum    186.88  exceeding 
the  whole  by  23.04  feet. 

Measured  as  the  Frustums  of  Cones; 
The  whole  tree  measures         193.53856  feet. 


The  great  end  measures  157.88672  feet. 

The  less  end  measures  35.65184  feet. 


Sum  193.53856  equal  to 
the  whole,  as  it  ought.  U 


218  Tne  Mensuration  of  Timber. 

2.  Dr.  Button,  p.  610,  Prob.  V. 

Supposing  a  tn>e  to  girt  11  Reel  at  the  greater  end, 
2  feet  at  the  lest,  and  its  length  to  be  83  Feel ;  where 
must  it  be  cut  that  the  part  next  the  greater  end  may 
measure  to  more  than  the  whole,  by  the  customary 
measure  ? 

RULE. 


Cut  it  through  (if  possible)  where  the  girt  is  4  of  the 

greatest.     This,  according  to  Dr.  Ilutton's  rules,  will 

be  7-£  feet  from   the    less  end,   and  24-|  feet  from  the 

greater  end  ;  and  the  girt  at  the  section  will  be  \4  or 

;et: 

By  the  common  Method. 

The  greater  end  measures     -     135£i  feet. 
The  whole  tree  measures     -     -128     feet. 


Diff.  7*\  so  that  the 

jpart  exceeds  the  whole  by  7  feit. 

By  Br.  Hutton's  Rule. 

The  greater  end  measures     -     173.44  feet. 
The  whole  tree  measures     -       163.84  feet. 


Diff*.  9.60  so  that  the 

part  exceeds  the  whole  by  9.6  feet. 

3.  Dr.   Ilutton,  p.  617,  Prob.  VI. 

Supposing  a  tree  to  girt  14  feet  at  the  greater  end, 
2  feet  at  the  less,  and  its  length  to  be  32  feet :  where 
must  it  be  cut  that  the  part  next  the  greater  end  may 
measure  exactly  to  the  same  as  the  whole,  by  the  cus- 
tomary method  ? 

By  Dr.  Huttonh  rules  the  lengths  of  the  two  parts 
must  be  13.599118  feet,  and  18.100882  feet ;  also  the 
girt  at  the  section  must  be  7.099669. 


The  Mensuration  of  Timber.  219 

By  the  common  Method. 

The  greater  end  measures         -        -        128  feet. 
The  whole  tree  measures  -         -         128  feet. 

Notwithstanding  above  ^  part  is  cut  off  the  length. 

By  Dr.  Huttotfs  Rule. 

The  greater  end  measures         -         163.8399  feet, 
The  whole  tree  measures         -     -      163.84     feet, 
notwithstanding  above  -}  part  is  cut  off  the  length. 

The  following  rules  for  measuring  timber  are  very 
accurate,  provided  the  dimensions  can  be  truly  taken, 
and  that  the  timber  be  in  the  form  of  a  parailelopi- 
pedon,  cylinder,  frustum  of  a  rectangular  pyramid,  or 
the  frustum  of  a  cone  ;  but  for  the  reasons  already 
given,  it  is  not  probable  that  they  will  ever  be  brought 
into  general  use. 

PROBLEM    I. 

To  find  the  Solidity  of  Timber  Scantling,  or  Squared 
Timber,  being  of  equal  breadth  aud  thickness  through- 
mit. 

RULE. 

Multiply  the  breadth  by  the  thickness,  and  that  pro- 
duct by  the  length. 

Thus,  you  will  find  the  answer  to  the  first  exam- 
ple, p.  211,  to  be  72.93  feet,  or  72  feet  11  inches. 

PROBLEM    II. 

To  find  the  Solidity  of  a  piece  of  Timber,  supposing  it 
to  be  PERFEcfLr  cylindrical. 

RULE. 

Multiply  the  square  of  one-fourth  of  the  circumfe- 
rence by  the  length  :  then  say,  as  11  is  to  14,  so  is  this 
content,  to  the  true  content. — Or,  use  Dr.  Hutton's 
rule,  see  page  26 1 . 


C20  Tlie  Mensuration  of  Timber. 

Thus,  yon  wiil  find  the  answer  to  the  5th  example, 
rage  Jl),  to  bey  1.(53  feet;  and  to  the  6th  example 
•1. IS  feet 

PROBLEM    III. 

To  find  the  Solidity  of  Squared  Timber,  tapering 
regularly, 

RULE. 

Multiply  the  breadth  at  each  end  by  the  depth,  and 
also  the  sum  of  the  breadths  by  the  sum  of  the  depth* 
These  three  products  added  together*  and  the  sum  mul- 
tiplied by  one-sixth  of  the  length,  will  give  the  so- 
lidity. 

Thus,  you  will  find  the  answer  to  the  3d  example, 
page  212,  to  be  43.101  feet,  and  to  the  4th  example 
37.33. 

To  find  the  Solidity  of  round  tapering  Timber,  hav- 
ing the  Girt  or  Circumference  of  the  two  ends  given 
in  inches,  and  the  length  in  feet. 

RULE. 

To  the  squares  of  the  two  circumferences  add  the 
square  of  their  sum;  multiply  this  sum  by,  the  length, 
cut  off  four  figures  from  the  right  hand  for  decimals, 
or  move  the  decimal  point  four  places  to  the  left  hand, 
and  J  |  of  the  product  will  be  the  content. 

Thus,  you  will  find  the  answer  to  the  7th  example, 
page  213,  to  be  74.38  feet,  and  the  8th  example  92.54 
feet. 

Note.  By  taking  the  first  example  in  the  scholium, 
page  217,  you  will  find 

The  content  of  the  whole  tree     193.931  feet. 


The  content  of  the  greater  end      158.231  feet. 
The  content  of  the  less  end  35.729  feet. 


Sum  193.960  equal  to 

the  whole  very  nearly. 


Tiie  Mensuration  of  Carpenters'  Work,  #c.   221 


CHAPTER  IV. 


Of  Measuring  the  Works  of  the  several  Artificers 
relating  to  Building  ;  and  what  Methods  and  Cus- 
toms are  observed  in  doing  it 


§  I.  Of  Carpenters'  and  JotitsRs9  Work. 

THE  Carpenters'  and  Joiners'  works,  which  are  mea* 
surable,  are  flooring,  partitioning,  roofing,  wainscot- 
ing, &c. 

I.  Of  Flooring. 

Joists  are  measured  by  multiplying  their  breadth  by 
their  depth,  and  that  product  by  their  length.  They 
receive  various  names  according  to  the  position  in 
which  they  are  laid  to  form  a  floor ;  such  as  trimming 
joists,  common  joists,  girders,  binding  joists,  bridging 
joists,  and  cieling  joists. 

In  boarded  flooring,  the  dimensions  must  be  taken  to 
the  extreme  parts,  and  the  number  of  squares  of  100 
feet  must  be  calculated  from  these  dimensions.  De- 
ductions must  be  made  for  stair-cases,  chimneys,  &c. 

EXAMPLES. 

1.  If  a  floor  be  57  feet  3  inches  long,  and  28  feet  6 
inches  broad ;  how  many  squares  of  flooring  are  there 
in  that  room  ? 


77/e  Mensuration  of  Carpenters'  Work,  #c. 

ft y  Decimals.         By  Duodecimals, 
tf.25  F.      1. 

57  :  6 

28  :  -; 


28625 
40800 
11450 


450 

114 

28 

7  : 

6 

7 

0  : 

0 

100  I  1631.625  feet. 

Squares       16.31625  16  I  31   :  7  :  6     , 

Facit  16  squares   and    31  feet. 

2.  Let  a  floor  be  53  feet  6  inches  long,  and  47  feet 
9  inches  broad;  how  many  squares  are  contained  in 
that  floor  ?  Jlns.  25  squares  54  feet. 

3.  A  floor  being  36  feet  3  inches  long,  and  16  feet 
6  inches  broad;  flow  many  squares  are  contained  in 
it  ?  Jlns.  5  squares  98  feet. 

4.  In  a  naked  floor  the  girder  is  1  foot  2  inches  deep, 
A  foot  broad,  and  20  feet  long;  there  are  8  bridging 
joists,  whose  scantlings  (viz.  breadths  and  depths)  are 
&  inches  by  6|  inches,  and  20  feet  long;  8  binding 
joists,  whose  lengths  are  9  feet,  and  scantlings  8|  inches 
by  4  inches  :  the  cieling  joists  are  24  in  number,  each 
6  feet  long,  and  their  scantlings  4  inches  by  2\  inches. 
Required  the  solidity  of  the  whole?  Jlns.  72  feet. 

5.  Suppose  a  house  of  three  stories,  besides  the 
ground  floor,  was  to  be  floored  at  £.6  10s.  per  square  ; 
♦he  house  measures  20  feet  8  inches,  by  16  feet  9 
inches ;  there  are  7  tire-places,  the  measures  whereof 
are,  two,  each  of  6  feet  by  4  feet  6  inches ;  two  other, 
each  of  6  feet  by  5  feet  4  inches ;  and  two,  each  of 
5  feet  8  inches  by  4  feet  8  inches  ;  and  the  seventh, 
5  feet  2  inches  by  4  feet.  The  well-hole  for  the  Stairs 
is  10  feet  6  inches  by  8  feet  9  inches.  What  will  the 
whole  tome  to  ?  Ms.  1.53.  13s.  3£& 


Tfie  Mensuration  of  Carpenters9  Work,  <$*c.     25& 


II.  Of  Partitioning. 

Boarded  partitions  are  measured  by  the  square  in 
the  same  manner  as  flooring,  and  deductions  must  be 
made  for  doors  and  windows,  except  they  are  included 
by  agreement. 

The  strongest  partitions  are  those  made  with  framed 
timber,  all  the  parts  of  whieh  are  measured  as  in 
flooring,  except  the  king-posts,  and  these  are  of  the 
same  kind  as  in  roofing,  an  example  of  which  will  be 
given  in  the  next  article. 


EXAMPLES. 

1.  If  a  partition  between  rooms  be  in  length  82  feet 
6  inches,  and  in  height  12  feet  3  inches  ;  how  many 
squares  are  contained  therein  ? 

Jlns.  10  squares  10  feefc 

2.  If  a  partition  between  rooms  be  in  length  91  feet 
9  inches,  and  its  breadth  11  feet  3  inches;  how  many 
squares  are  contained  in  it  ? 

Jlns.  lo  squares  32  feefc 

III.    Of  Roofing. 

In  roofing  take  the  whole  length  of  the  timber 
for  the  length  of  the  framing,  and  for  the  breadth  gird 
over  the  ridge  from  wall  to  wall  with  a  string.  This 
length  and  breadth  must  be  multiplied  together  for  the 
content. 

It  is  a  rule  also  among  workmen,  that  the  flat  of 
any  house,  and  half  the  flat  thereof,  taken  within  the 
walls,  is  equal  to  the  measure  of  the  roof  of  the  same 
house;  but  this  is  when  the  roof  is  of  a  true  pitch. 
The  pitch  of  every  roof  ought  to  be  made  according 
to  its  covering,  which  in  England  is  lead,  pantiles, 
plaJHtiles,  or  slates.    The  usual  pitches,  are  the  pedf- 


Tlie,  Menswution  of  Carpenters'  Work,  $c. 

ment  pitch,  used  when  the  covering;  is  lead ;  the  per- 
pendicular height  is  -|  of  the  breadth  of  the  building. 
The  common,  or  true  pitch,  where  the  Length  of  the 
ratters  are  }  of  the  bread."  of  the  building;  this  is 
used  when  the  covering  is  plainliles.  The  Gothic 
pitch,  is  when  the  length  of  the  principal  rafters  is 
equal  to  the  breadth  of  the  building,  forming  an  equi- 
lateral triangle :  This  pitch  is  used  when  the  covering 
is  of  pantiles. 

In  the  measuring  of  roofing  for  workmanship  alone, 
all  holes  for  chimney-shafts  and  sky-lights  are  gene- 
rally deducted.  But  in  measuring  for  work  and  mate-  I 
rials,  they  commonly  measure  in  all  sky-lights,  luthern- 
lights,  and  holes  for  the  ehimney -shafts,  on  account  of  • 
their  trouble,  and  waste  of  materials. 


EXAMPLES. 

1.  If  a  house  within  the  walls  be  44  feet  6  inches' 
long,  and  18  feet  3  inches  broad;  how  many  square* 
«f  roofing  will  cover  that  house  ? 


By  Decimals. 

±8.25 

44.5 

ft 

The  flat 
JThe  half 

Sum  1 
Facit  12  squan 

1  Duodecimals. 
F.     I. 
44  :  6 
18  *  3 

9125 

t        7300 
7300 

352 

44 

11:1*6 

Flat     812.125 
Half    406. 

9:0:0 

812  :  1  :  6 

100  |  12    18 

406 

12.18 

L2|18 

es  18  fee*. 

The  Mensuration  of  Carpenters'  Work,  Sfc.     22& 

2.  What  <  i»<  the  roofing  of  a  house  at  los.  6d.  per 
square;  the  length,  within  the  walls,  being  52  feet 
8  inches,  and  the  breadth  30  feet  6  inches ;  the  roof 
being  of  a  true  pitch?  Ms. 1.12  12s.  llfd. 

3.  The  roof  of  a  house  is  of  a  true  pitch ;  and  the 
house  measures  40  feet  6  inches  in  length,  within  the 
walls,  and  20  feet  6  inches  in  breadth;  how  many 
squares  of  roofing  are  contained  therein  ?  * 

Ms.  12.45375. 
Note.  All  timbers  in  a  roof  are  measured  in  the 
same  manner  as  in  floors,  except  king-posts,  &c.  such 
as  A  in  the  annexed  figure,  where  there  is  a  necessity 
for  cutting  out  parallel  pieces  of  wood  from  the  sides, 
in  order  that  the  ends  of  the  braces  B,  that  come 
against  them,  may  have  what  the  workmen  call  a 
square  hutment.  To  measure  king-posts  for  workman- 
ship, take  their  breadth  and  depth  at  the  widesf 
place,  and  multiply  these  together,  and  the  product 
by  the  length.  To  find  the  quantity  of  materials,  if 
the  pieces  sawn  out 
are  2-  inches  broad, 
or  upwards,  and 
more   than     2  feet  S>* 

long,  they  are  con-  ^^\ 
sidered  as  pieees  of  f  ~^ 
timber  fit   for   use. 

In  measuring  these  pieces,  the  shortest  length  must 
always  be  taken,  because  the  sawing  of  them  from  the 
king-post,  renders  a  part  of  them  useless.  The  soli- 
dity of  these  must  be  deducted  from  the  solidity  of  the 
king-post.  "When  the  pieces  sawn  out  are  of  smaller 
dimensions  than  above  described,  the  whole  post  is 
measured  as  solid  without  any  deduction,  the  pieces 
cut  out  being  esteemed  of  little  or  no  value. 

4.  Let  the  tie-beam  D,  in  the  above  figure,  be  36 
feet  long.  9  inches  broad,  and  1  foot  2  inches  deep; 
the  king-post  All  feet  6  inches  high,  1  foot  broad 


Vie  Mensuration  of  Carpenters"  ffbrh,  <Jv. 

at  the  bottom,  and  0  inches  thick  :  out  of  this  post 
arc  sawn  two  equal  pieces  from  the  sides,  each  7  feet 
long  and  3  inches  broad.  The  braces  B,  B,  are  7 
feet  6  inches  long,  and  5  inches  by  5  inches  square: 
the  rafters  E,  B,  are  1<>  feet  long,  S  inches  broad,  and 
10  inches  deep:  the  struts,  C,  (',  are  3  feet  6  inches 
long,  4  inches  broad,  and  5  inches  deep.  Required 
#4 he  measurement  for  workmanship,  and  likewise  for 
materials  ? 


F.        I.     P. 

31  :     6:0  solidity  of  the  tie-beam  D. 

4  :     9:6  solidity  of  the  king-post  A. 

2  :  7:3  solidity  of  the  bracks  B,  B. 
13  :  2:4  solidity  of  the  rafters  E,  E. 
—  :  11  :  8  solidity  of  the  struts  C?  C. 


I 


53  :     0:9  solidity  for  workmanship. 
1  :     5:6  solidity  cut  from  the  king-post. 

51  :     7:3  solidity  for  materials. 


IV.     Of  Wainscoting,  Sfc. 

Wainscoting  is  measured  by  the  yard  square,  con- 
sisting of  9  square  feet.  The  dimensions  are  taken 
in  feet  and  inches  ;  thus,  in  taking  the  height  of  a 
room  they  girt  over  the  cornice,  swelling  pannels,  and 
mouldings,  with  a  string;  and  for  the  compass  of  the 
room,  they  measure  round  the  floor.  Doors,  window- 
shutters,  and  such  like,  where  both  sides  are  planed, 
are  considered  as  work  and  half;  therefore  in  mea- 
suring a  room  they  need  not  be  deducted,  but  the 
room  may  be  measured  as  if  there  were  none;  then 
the  contents  of  the  doors  and  shutters  must  be  found, 
and  the  half  thereof  added  to  the  content  of  the  whole 
room. 


Tlie  Mensuration  of  Carpenters'  Work,  $c.     %$2 

Windows,  where  there  are  no  shutters,  must  be  de- 
ducted; also  chimneys,  window-seats,  cheek-boards, 
sopheta-boards,  linings,  &c.  must  be  measured  by  them- 
selves. 

Weather-boarding  is  measured  by  the  yard  square^ 
and  sometimes  by  the  square. 

Windows  are  generally  made  and  valued  by  the 
foot  superficial  measure,  and  sometimes  at  so  much 
per  window.  When  they  are  measured,  the  dimen- 
sions must  be  taken  in  feet  and  inches,  from  the  under 
side  of  the  sill  to  the  upper  side  of  the  top-rail,  for 
the  height ;  and  for  the  breadth,  from  outside  to  out- 
*side  of  the  jambs.  This  length  multiplied  by  the 
breadth  will  be  the  superficies. 

Stair-cases  are  measured  by  the  foot  superficial, 
and  the  dimensions  are  taken  with  a  string,  girt 
over  the  riser,  and  tread ;  and  that  length,  or  girt, 
multiplied  by  the  length  of  the  step,  will  give  the  su- 
perficies. 

The  rail  is  taken  at  so  much  per  foot  in  length, 
according  to  the  diameter  of  the  well-hole;  archi- 
trave string-boards  by  the  foot  superficial  ;  brackets 
and  strings  at  so  much  per  piece,  according  to  the 
workmanship. 

Door-cases  are  measured  by  the  foot  superficial,  ami 
the  dimensions  must  be  taken  with  a  string  girt  round 
the  architrave  and  inside  of  the  jambs,  for  the  breadth; 
and  for  the  length,  add  the  length  of  the  two  jambs  to 
the  length  of  the  eap-pieee  (taking  the  breadth  of  the 
opening  for  the  length)  the  product  of  these  two  will 
be  the  superficies. 

Frame  doors  are  measured  by  the  foot,  or  sometimes 
by  the  yard  square. 

Modillion  cornices,  coves,  &e.  are  generally  measur- 
ed by  the  foot  superficial. 

Frontispieces  are  measured  and  valued  by  the  foot 
superficial,  and  every  part  is  measured  separately,  viz\ 
architrave,  frieze,  and  cornice. 


I 


229    The  Mensuration  of  Carpenters*  Work,  cjv. 


EXAMPLES. 


1.  If  a  room,  or  wainscot",    being   girt    downwards 
over  the  mouldings,  be.     u  feet  9  inches  high,  and  126 
inches  in  compass;  how  many  yards   does  that 
room  contain  ? 


By  Duodecimals. 
F.       I. 
126  :  3 
15  :  9 

By  Decimals. 

126.25 
15.75 

630 
126 
63  :  1  :  6 
31  :  6  :  9 
3:9:0 

63125 

88375 
63125 
12625 

9)1988.4375 

9)1988  :  5  :  3 

220.8 
Jlns.  220  yards  S  feel. 

Ms.  220.8 

2.  If  a  room  of  wainscot  he  16  feet  3  inches  high, 
and  the  compass  of  the  room  137  feet  6  inches;  how 
many  yards  are  contained  in  it  ? 

Jlns.  2-18  yards  2  feet. 

3.  If  the  window-shutters  about  a  room  he  69  feet  9 
inches  broad,  and  6  feet  3  inches  high,  how  many  yawls 
are  contained  therein  at  work  and  half  ? 

Jlns.  72.656  yards. 

4.  What  will  the  wainscoting  of  a  room  come  to  at 
6  shillings  per  square  yard,  supposing  the  height  of 
the  room,  including  the  cornice  and  moulding,  be  12 
feet  6  inches,  and  the  compass  83  feet  8  inches  ;  three 
window-shutters,  each  7  feet  8  inches  by  2  feet  6  inch- 
es, and  the  door  7  feet  by  3  feet  6  inches  ;  the  shutters 
and  door  being   worked  on  both  sides,  are   reckoned 


work  and  half  ? 


dns.  1.36  4s.  6|d. 


Jim 


The  Mensuration  of  Bricklayers'  Work,  <J*c.    22% 

5.  A  rectangular  room  measures  129  feet  5  inches 
round,  and  is  to  be!  wainscoted  at  3s.  f>d.  per  square 
yard.  After  dua  allowance  for  girt  of  cornice,  &c. 
it  is  16  feet  3  inches  high;  the  door  is  7  feet  by  3  feet 
9  inches:  the  window  shutters,  two  pair  ai -  7  feet  3 
inches  by  4  fret  6  inches ;  the  cheek-boards  round  ihem 
come  13  inches  below  the  shutters,  and  are   14  inches 

breadth;  the  fining-boards  round  the  door-way  are 
16  inches  broad;  the  door  and  window-shutters  being 
worked  on  both  sides,  are  reckoned  as  work  and  half, 
and  paid  for  accordingly  :  the  chimney  3  feet  9  inches 
by  3  feet,  not  being  enclosed,  is  to  be  deducted  from 
the  superficial  content  of  the  room.  The  estimate  of 
the  charge  is  required  ?  Jlns.  Z.43  :4s.  :6fd. 


§  II.  Of  BmcKLArERS'  Work. 
The  principal  is  tiling,  walling,  and  chimney -work, 

I.  Of  Tiling. 

Tiling  is  measured  by  the  square  of  100  feet,  as 
.flooring,  partitioning,  and  roofing  were  in  the  carpen- 
ter's work ;  so  that  between  the  roofing  and  tiling,  the 
difference  will  not  be  much ;  yet  the  tiling  will  be  the 
most;  for  the  bricklayers  sometimes  will  require  to 
have  double  measure  for  hips  and  vallies.  When  gut- 
ters are  allowed  double  measure,  the  way  is  to  measure 
the  length  along  the  ridge-tile,  and  by  that  means  the 
measure  of  the  gutters  becomes  double :  it  is  usual  al- 
so to  allow  double  measure  at  the  eaves,  so  much  as  the 
projection  is  over  the  plate,  which  is  commonly  about 
IS  or  20  inches. 


\v 


iration  of  Bricklayers'  Work,  $c. 


mi-i.i.s. 

I  ii<  iv  i>  ;i  roof  covered  willi  tiles,  whose  depth 
on  both  sides  (v  itli  (lie  usual  allowance  at  the  eaves) 
is  .;:  feet  -J  inches,  and  the  length  43  feet  ;  1  demand 
how  many  squares  of  til  jug  are  contained  therein? 

By  Duodecimals.  By  Decimals. 

\\     I.  87,28 

;}7  :  3  45 

13  :  0  


18623 

1S3  11900 

^L4S 


11. :  3  16  |  76.23 


16176  :  3 

Jns.  16  squares,  76  feet. 

-.  There  is  a  roof  covered  with  tiles,  whose  depth 
yii  both  sides  (with  the  allowance  at  the  eaves)  is  35 
feet" 9  inches,  and  the  length  43  feet  6  inches;  I  de- 
mand haw  many  squares  of  tiling  are  in  the  roof? 

Jln§.  13  squares  55  feet. 

3.  What  will  the  tiling  a  barn  cost  at  lA  5s.  6d.  per 
square,  the  length  being  43  feet  10  inches,  and  breadth 
27  feet  5  inches,  on  the  Hat,  the  eaves  projecting  16 
inches  on  eacli  side  ?  Jlns.  1,24  9s.  5^d. 


2' 


IT.     Of  Walling. 

Bricklayers  commonly  measure  their  work  by  the 
rod  square  of  16  feet  and  an  half;  so  thai  one  rod  in 
length,  and  one  in  breadth,  contain  272.25  square  feel ; 
for  f1i.5,  multiplied  into  itself,  produces  272.23  square 
feet.     But  in  some  places  the  custom  is  to  allow    18 


The  Mensuration  of  Bricklayers'  Work,  Sfc.    231 

feet  to  the  rod  ;  that  is,  321  square  feet.  And  in 
some  places  the  Usual  way  is,  to  measui-6  by  the  rod  of 
2t  feet  long  and  3  feet  high,  that  is,  63  square  feet ; 
and  here  they  never  regard  the  thickness  of  the  wall 
in  measuring,  but  regulate  the  price  according  to  the 
thickness. 

When  you  measure  a  piece  of  brick-work,  the  first 
thing  is  to  enquire  by  which  of  these  ways  it  must  be 
measured,  then,  having  multiplied  the  length  and 
breadth  in  feet  together,  divide  the  product  by  the  pro- 
per divisor,  viz.  272.23 ;  324  ;  or  63  ;  according  to 
the  measure  of  the  rod>  and  the  quotient  will  be  the 
answer  in  square  rods  of  that  measure. 

But  commonly  brick-walls,  that  are  measured  by  the 
rod,  are  to  be  reduced  to  a  standard  thickness  ;  viz.  of 
a  brick  and  a  half  thick  (if  it  be  not  agreed  on  the 
contrary;)  and  to  reduce  a  wall  to  standard  thick- 
ness, this  is 

THE    RULE. 

Multiply  the  number  of  superficial  feet  that  are 
found  to  be  contained  in  any  wall  by  the  number  of 
half-bricks  which  that  wall  is  in  thickness  ;  one-third 
part  of  that  product  shall  be  the  content  in  feet,  re- 
duced to  the  standard  thickness  of  one  brick  and  a 
half. 

EXAMPLES. 


1.  If  a  wall  be  72  feet  6  inches  long,  and  19  feet  3 
inches  high,  and  5  bricks  and  a  half  thick  ;  how  ma- 
ny rods  of  brick-work  are  contained  therein,  when  re- 
duced to  the  standard  ? 


The  Mensuration  of  Bricklayers-  Work,  tj*c. 


By  Decimals. 
I 

9625 
3850 


1305.625 
11 

:)  153  5 1.875 


.2:2.25)5117.291(18  rods. 
239479 
63.06)216.79(3  quar. 


12.61  feet. 


Bit  Duodecimals. 
V.      J. 
72  :   6 
19  :  3 


648 

IS  :  1  :  6 
9    :  6  :  - 


1395 

:  7 

:  6 
11 

3)15351 

10 

:  6 

272)5117  : 
2397 

(18 
quar 

rods 

68)221(3 

ters. 

feel. 


Note.  That  68.06  is  one -fourth  part  of  272.25, 
aud  68  is  one-fourth  of  272. 

In  reducing  feet  into  rods,  it  is  usual  to  reject  the 
odd  parts,  and  divide  only  by  272,  as  is  done  in  the 
second  way  of  the  last  example  ;  here  the  auswer  is 
18  rods  3  quarters  and  17  feet ;  about  4£  feet  more 
than  by  the  first  way,  where  it  is  done  decimally;  a 
difference  too  trilling  to  be  considered  in  practice. 

To  find  proper  Divisors  for  bringing  the  Answer  in 
feet,  or  rods,  of  the    Standard  thickness;    without 
multiplying  the   Superficies,  by  the   number  of  half- 
bricks,  <$*c. 

Divide  3,  the  number  of  half-bricks  in  ll2,  by  the 
number  of  half-bricks  in  the  thickness,  the  quotient 
will  be  a  divisor,  which  will  give  the  answer  in  feet. 


The  Mensuration  of  Bricklayers'  Work,  Sfc.    23  3 


But  if  you  would  have  a  divisor  to  bring  the  rnswer 
in  rods  at  once,  multiply  272  by  the  divisor  found  for 
feet,  am?  the  product  will  be  a  divisor  for  rods  ;  as  in 


the  following  table. 


Divisors 

Divisors  for 

The  thickness 

for  the 

the  answer 

of  the  wall. 

answer 
in  feet. 

in  rods. 

1     Brick 

1.5 

408. 

1§  Brick 

1. 

272. 

2    Bricks 

.75 

204. 

2±  Bricks 

.6 

163.2 

3     Bricks 

.5 

136. 

3f  Bricks 

.4285 

116.6 

4     Bricks 

.373 

102. 

5\  Bricks 

.2727 

74.17 

By  Scale  and  Compasses. 

Extend  the  compasses  from  the  tabular  divisor 
against  the  given  thickness,  to  the  length  of  the  wall, 
that  extent  will  reach  from  the  breadth  to  the  con- 
tent. 

Or  by  the  Carpenter' 's  Rule, 

As  the  tabular  divisor,  against  the  thiekness  of  the 
wall,  is  to  the  length  of  the  wall ;  so  is  the  breadth  to 
the  content. 

Taking  the  preceding  example,  extend  the  eompas 


ses  from  74.17  to 
19.23  tol8|rods. 


'2.5.  that   extent  will  reach   from 


By  the  Carpenter's  Mule. 


B         A 

74.17  :  72.5 


B 

:  19.23 


A 

181 


The  Mensuration  of  Bricklayers*  Work,  $ei 


The  dimensions  of  b  building  arc  generally  taken 
by  measuring  half  round  the  outside,  and  half  round 
the  inside,  for  the  whole  length  of  the  wall;  this 
length  being  multiplied  by  the  height  gives  the  super- 
ficies. And  to  reduce  it  to  the  standard  thickness,  8a 
proceed  ns  above.  All  the  vacuities,  such  as  doors, 
windows,  window-backs,  &c.  must  be  deducted. 

To  measure  an  arched-way,  arched-window,  or 
doors,  &c.  take  me  height  of  the  window,  or  door, 
from  the  crown  or  middle  of  the  arch,  to  the  bottom 
or  sill  ;  and  likewise  from  the  bottom  or  sill  to  the 
spring  of  the  arch,  that  is,  where  the  arch  begins  to 
turn.  Then  to  the  latter  height  add  twice  tlie  for- 
mer, and  multiply  the  sum  by  the  width  of  the  win- 
dow, door,  &c.  and  one-third  of  the  product  will  be 
the  area,  sufficiently  near  for  practice. 

2.  If  a  wall  be  245  feet  9  inches  long,  16  ftet  6 
inches  high,  and  two  bricks  and  a  half  thick  ;  how 
many  rods  of  brick-work  are  contained  therein,  when 
reduced  to  the  standard  thickness. 

dns.  24  rods  3  quarters  24  feet. 

3.  How  many  rods  are  contained  in  a  wall  63|  feet 
long,  14  feet  11  inches  high,  and  2£_  bricks  in  thick- 
ness, when  reduced  to  the  standard  ? 

Jlns.  5  rods  218  feet. 

4.  A  triangle  gable-end  is  raised  to  the  height  of 
15  feet  abo\e  the  end-wall  of  a  house,  whose  width 
is  45  feet,  and  the  thickness  of  the  wail  is  2\  bricks; 
required  the  content  in  rods  at  standard  thickness  ? 

Jlns.  2  rods  18  feet. 

5.  Admit  the  end-wall  of  a  house  to  be  28  feet  10 
inches  in  breadth,  and  the  height  of  the  roof  from  the 


The  Mensuration  of  Bricklayers'  Work,  $c.     233 

ground  55  feet  8  inches,  the  gable  (or  triangular  part 
above  the  side  walls)  to  rise  42  courses  of  bricks,  reck- 
oning 4  courses  to  a  foot ;  and  that  20  feet  high  be  2\ 
bricks  thick,  20  feet  more  2  bricks  thick,  and  the  re- 
maining 15  feet  8  inches  t£  brick  thick ;  what  will 
the  work  come  to  at  1.5  10s.  per  rod,  the  gable  being 
pne  brick  in  thickness  ?  Jins.  £.48  13s.  5£d, 


III.     Of  Chimneys. 

If  you  are  to  measure  a  chimney  standing  by  itself,. 
without  any  party-wall  being  adjoining,  then  girt  it 
about  for  the  length,  and  the  height  of  the  story  is 
the  breadth ;  the  thickness  must  be  the  same  as  the 
jambs  are  of,  provided  that  the  chimney  be  wrought 
upright  from  the  mantle-tree  to  the  cieling,  not  deduct- 
ing any  thing  for  the  vacancy  between  the  floor  (or 
hearth)  and  the  mantle-tree,  because  of  the  gatherings 
of  the  breast  and  wings,  to  make  room  for  the  hearth 
in  the  next  story. 

\i  the  chimney-back  be  a  party-wall,  and  the  wall 
be  measured  by  itself,  then  you  must  measure  the 
depth  of  the  two  jambs,  and  the  length  of  the  breast 
for  a  length,  and  the  height  of  the  story  for  the  breadth, 
at  the  same  thickness  your  jambs  were  of. . 

When  you  measure  chimney-shafts,  viz.  that  part 
which  appears  above  the  roof,  girt  them  with  a  line 
round  about  the  least  place  of  them,  for  the  length, 
and  take  the  height  for  the  breadth :  and  if  they  be 
four  inches  thick,  then  you  must  set  down  their  thick* 
ness  at  one  brick-work ;  but  if  they  be  wrought  9 
inches  thick  (as  sometimes  they  are,  when  they  stand 
high  and  alone  above  the  roof,)  then  you  must  account 
your  thickness  i\  brick,  in  consideration  of  plastering? 
and  the  trouble  in  scaffolding. 


novation  of  Bricklayers'  Work,  <jjv. 

If    is  customary,    in  most  places,  to   allow  double 
me  foe  chimneys. 


-r-   r    r 

J  ;     ■  '  m 

I 


1.  Suppose  the  figure,  ABCDEFGHIK,  to  be  a 
chimney  that  hath  a  double  funnel  towards  the  top, 
and  a  double  shaft,  ABC.  Then  suppose  in  the  par- 
lour, the  breast  IL  and  the  two  jambs  HI  and  LK 
to  measure  13  feet  9  inches,  and  the  height  of  the  room 
HF,  or  Irt,  to  be  12  feet  6  inches ; — in  the  first  floor 
let  the  breast  and  the  two  jambs,  viz.  ab,  girt  14  feet 
6  inches,  and  the"  height  «1),  be  9  feet  5 — in  the  second 


The  Mensuration  of  Bricklayers9  Work,  Sec    237 

floor  let  the  breast  and  the  jambs,  viz.  cc?,  girt  10  feet  3 
inches,  and  the  height  cB  be  7  feet; — above  the  roof, 
let  the  compass  of  the  shaft  ABC  be  13  feet  9  inches, 
and  its  height  AB  6  feet  6  inches ;  lastly,  let  the  length 
of  the  middle  partition,  which  parts  the  funnels,  be  12 
i'eet,  and  its  thickness  1  foot  3  inches :  how  many  rods 
of  brick-work,  standard  measure,  are  contained  in  the 
chimney,  allowing  double  measure  ? 


F. 

1.     is 
12 

I. 

:  9 
:  6 

Fee 
Feet 

F.    I. 

1  :  3 

±2  :  - 

225 
9 

:  0 
:  4  :  6 

15  :  0  partition. 

F.      I.   Pts. 

234  :  4  :  6  parlour. 
130  :  6  :   -  first  floor. 

71  :  9  :   -  sec.  floor. 

89  :  4  :  6  shaft. 

15  :  -  :   -  partition. 

FGHK234 

:  4  :  6 

P. 

2.     14 

9 

I. 

•  6 

Da  E6  130 

:  6 

t  541  :  -  :  -  sum. 

F. 

3       10 

I. 

:  3 

1082  :  -  :  -  double. 

272)1082(3     rods. 

Be   Cd  71 

:  9 

68)266(3     quarters. 

F. 

4.      13 
6 

I. 

:  9 
•  0 

62        feet. 

Ans.  3  rods,  3  quarters, 
and  62  feet,  admitting 
the  common  thickness 
to  be  li  briek. 

1 

:  6 

:  10  :  6 

ABC  S9 

:  4     :  6 

Note.     This  chimney  being  measured  as  if  it  were 
solid,  no  vacuity,  or   opening,   for  the  fire-place,  in 


238      Tlie  Mensuration  of  Plasterers'  Work,  §*c.> 

any  of  (lie  floors,  is  drawn  in  the  above  figure.  Also 
(he  jambs  HIKw,  Jik(«/>.  are  supposed  to  he  perpen- 
dicular to  the  breast  1L,  viz.  IUL  and  ILK  are  right 
angle* ;  the  tame  must  be  observed  in  the  other  two 
floors. 

2.  Suppose  the  breadth  TL  of  the  breast  of  a  chim- 
ney, not  standing  in  an  angle,  to  be  7  feet  :;  inches, 
the  depth  III  or  LIv  of  the  jambs  3  bricks  thick; 
the  height  of  the  room  Irt  14  feet  6  incites ;  the  vacuity 
for  the  tire-place  4  feet  high,  3  feet  6  inches  wide, 
and  3  bricks  deep.  The  chimney  and  fire-place  in 
each  of  the  other  two  rooms  are  of  the  same  dimen- 
sions, but  the  height  of  the  room  aD  is  12  feet;  and 
the  jamb  cB,  which  is  the  same  height  as  the  story, 
10  feet  6  inches.  The  shaft  BA  stands  6  feet  above 
the  roof,  and  its  compass  is  10  feet;  also  its  thickness 
is  estimated  at  !*■  brick;  how  many  rods  of  brick- 
work standard  thickness,  allowing  double  measure, 
are  contained  in  the  chimney,  deducting  the  three  fire- 
places. *2ns.  3  rods  3  quarters  5  feet. 


§111  Of  Plasterers'  Work. 

The  plasterers'  works  are  principally  of  two  kinds  \ 
namely,  1.  plastering  upon  laths,  called  cieling,  and 
2.  plastering  upon  walls,  or  partitions  made  of  framed 
timber,  called  rendering.  In  plastering  upon  walls, 
no  deductions  are  made  except  for  doors,  windows, 
&c.  but  in  plastering  timber  partitions,  in  large  ware- 
houses, &e.  where  several  of  the  braces  and  larger 
timbers  project  from  the  plastering,  a  fifth  part  is 
commonly  deducted.  Plastering  between  these  timbers 
is  generally  called  rendering  between  quarters. 

Whitening  and  colouring  are  measured  in  the  same 
manner  as  plastering  :  as  in  rendering  between  these 
projecting  timbers  one  fifth-part  is  deducted,  so  in  gu- 


The  Mensuration  of  Plasterers9  Work,  Sfc.    239 

louring,  it  will  be  necessary  to  add  one-fourth  or 
one-fifth  of  the  whole,  for  the  trouble  of  colouring  the 
projections. 

Plasterers'  work  is  measured  by  the  yard  square, 
consisting  of  9  square  feet.  In  arches,  the  girt 
round  them,  multiplied  by  the  length,  will  give  the 
superficies. 

EXAMPLES. 

1.  If  a  cieling  be  59  feet  9  inches  long,  and  24  feet 
<5  inches  broad;  how  many^ards  does  that  cieling 
contain  ? 


By  Decimals. 

59.75 

21.5 


By  Duodecimals. 

F. 

I. 

59 

:  9 

24 

:  6 

/ 

236 

118 

29 

:  10 

:  6 

18 

:     0 

:  0 

9)1-163 

:  10 

:  6 

29873 
23900 
11950 


9)1463.873  feet. 
:s.  162.65  yards. 


162  yards  5  feet. 


2.  If  the  partitions  between  rooms  be  141  feet  6 
inches  about,  and  11  feet  3  inches  high;  how  many 
yards  are  in  those  partitions  ?  Jins.  176.87  yards. 

3.  AVhat  will  be  the  expence  of  plastering  a  ciel- 
ing, at  llj-d.  per  yard,  supposing  the  length  22  feet  7 
inches,  and  breadth  13  feet  11  inches  ? 

Ms.  I A  13s.  0fd. 

4.  There  is  a  partition  which  measures  234  feet  8 
inches  round,  and  14  feet  6  inches  high,  this  partition 
is  rendered  between  quarters,  that  is,  it  is  made  of 
framed  timber,  and  the  interstices   are  filled  up  with 


960  The  Mensuration  of  Painters'  Work. 

hit h  and  plastering ,  The  lathing  and  plastering  wiH 
be  Bd.  per  yard,  and  the  whitening  2d.  per  yard;  what 
will  the  whole  eome  to  ? 

Ans.  j.18  IV*.   J'd. 

5.  The  length  of  a  room  is  14  feet  .1  inches,  breadth 
IS  feet  C  inches,  and  height 9  feet  •>  inches,  to  the  un- 
der side  of  the  cornice,  which  projects  5  inches  from 
the  wall,  on  the  upper  part  next  the  cieling  ;  required 
the  quantity  of  rendering  and  plastering,  there  being 
no  deductions  but  for  one  door,  the  size  whereof  is  7 
feet  by  i  .: 

Ans.  5o  ya#*ls  3  feet  of  rendering,  and  18 
yard*  3  feet  of  cie'ing. 

6.  The  circular  vaulted  iyf  of  a  church  measures  103 
feet  6  inches  in  the  arch,  "and  273  feet  5  inches  in 
length  5  what  will  the  plastef  ing  come  to  at  is.  per 
yard?  Ms.  U5Q  17s.  B\d. 

U/ 


§  IV.     Of  Paint-ers^  Work. 

The  taking  the  dimensions  of  painters'  work  is  the 
same  as  that  of  joiners,  by  girting  over  the  mouldings 
and  swelling  pannels  in  taking  the  height;  and  it  is 
but  reasonable  that  they  should  be  paid  for  that  on 
which  their  time  and  colour  are  both  expended.  The 
dimensions  thus  taken,  the  casting  up,  and  reducing 
feet  into  yards,  is  altogether  the  same  as  the  joiners' 
work,  but  the  painter  never  requires  work  and  half, 
but  reckons  his  work  once,  twice,  or  thrice  coloured 
over.  Only  take  notice,  that  window-lights,  window- 
bars,  casements,  and  such-like  things,  they  do  at  so 
much  a  piece. 

EXAMPLES. 

1.  If  a  room  be  painted,  whose  height  (being  girt 
over  the  mouldings)  is  10  feet  6  inches,  and  the  com- 


Tlie  Mensuration  of  Painters'  Work. 


Jll 


pass  of  the  room  97  feet  9  inches ;  how  many  yards  are 
in  that  room  ? 


By  Duodec 

F.    I. 

97  :  9 
16  :  6 

imals. 

:  6 

By  Decimals. 

97.5 
16.5 

584 

98 

48     :  10 

4S875 
58630 

'     9775 

1612  :  10 

6- 

9)1612.875 

179  :     1  179.2. 

Facit  179  yards  2  feet,  nearly. 

2.  A  gentleman  had  a  room  painted  at  S^d.  per  yard, 
the  measure  whereof  is  as  follows*;  the  height  11  feet 
7  inches,  the  compass  74  feet  10  inches,  the  door  7 
feet  6  inches  by  3  feet  9  inches  ;  five  window-shutters, 
each  6  feet  8  inches  by  3  feet  4  inches,  the  breaks  in 
the  windows  14  inches  deep  and  8  feet  high  ;  the  open- 
ing for  the  chimney  6  feet  9  inches  by  5  feet,  to  be  de- 
ducted; the  shutters  and  doors  are  coloured  on  both 
sides  ;  what  will  the  whole  come  to  ? 

Ans.  lA  6s.  lid. 

3.  Suppose  a  room  were  to  be  painted,  and  that  its 
length  is  24  feet  6  inches,  breadth  16  feet  3  inches, 
and  height  12  feet  9  inches  ;  also  the  size  of  the  door 
7  feet  by  3  feet  6  inches,  and  the  size  of  the  window- 
shutters  to  each  of  the  windows,  there  being  two,  is  7 
feet  9  inches  by  3  feet  6  inches ;  but  the  breaks  of  the 
windows  themselves  are  8  feet  6  inches  high,  and  1 
foot  3  inches  deep ;  what  will  be  the  expence  of  giving 
it 3  coats,  at  2d.  per  yard  each;  the  size  of  the  fire- 
place to  be  deducted,  being  5  feet  by  5  feet  6  inches  ? 

Jins.  1,3  3s.  104-. 
X 


143  The  Mensuration  of  Glaziers'  Work. 

§  V.     Of  Glaziers'  Work: 

Glaziers  take  their  dimensions  in  feet,  inches,  and 
eighths  or  tenths,  or  else  in  feet  and  hundredth  parts  of 
a  foot,  and  estimate  their  work  by  the  square-foot 

Windows  are  sometimes  m  asored  by  taking  the  di- 
mensions of  one  pane,  and  multiplying  its  superficies 
hv  the  nnniher  of  panes.  But  more  generally,  they 
measure  the  length  and  breadth  of  the  window  over  all 
the  pan  s  and  their  frames,  for  the  length  and  breadth 
of  the  glazing. 

Circular  or  oval  windows,  as  fan  lights,  See.  are 
measured  as  if  they  were  square,  taking  for  their  di- 
mensions the  greatest  length  and  breadth  ;  as  a  com- 
pensation for  the  waste  of  glass,  and  labour  in  cutting 
it  to  the  nec/ssary  forms. 

Plumbers'  work  is  rated  at  so  much  per  pound,  or  by 
the  hundred-weight  of  112  pounds. 

Sheet-lead,  used  in  rooting,  guttering,  &c.  weighs 
from  7  to  12  pound  per  square  foot,  according  to  the 
thickuess.  And  a  pipe  of  an  inch  hore  weighs  com- 
monly 13  or  14  pounds  per  yard  in  length. 

EXAMPLES. 

1.  If  a  pane  of  glass  he  4  feet  8  inches  and  3  quar- 
ters long,  and  1  foot  4  inches  and  1  quarter  broad  : 
how  many  feet  of  glass  are  in  that  pane  ? 

By  Duodecimals.  By  Decimals. 

ft    I.     P.  4.729 

4:8:9  1.334 

1:4:3  


18916 

4:8:9  21645 

1  :  v6  :11  :  0  14187 

1:2:2:3  4729 


6  :  4  :10  :  2  :  3  6.403066 

Ms.  6  feet  4  inches. 


The  Mensuration  of  Glaziers9  Work.         243 

2.  If  there  be  8  panes  of  glass  eaeh  4  feet  7  inches 

3  quartos  long,  and  1  foots  inches  1  quarter  broad; 
how  many  feet  of  glass  are  contained  in  the  said  8 
panes  ?  Ms.  53  feet  5  inches. 

3.  If  there  be  16  panes  of  glass,  each  4  feet  5  inch- 
es and  a  half  long,  and  1  foot  4  inches  3  quarters 
broad;  how  many  feet  of  glass  are  contained  in  them? 

Jlns..  99  feet  6  inches. 

4.  If  a  window  be  7  feet  3  inches  high,  and  3  feet 
B  inches  broad  ;  how  many  square  feet  of  glazing  are 
contained  therein  ?  Jlns.  24  feet  9  inches. 

5.  There  is  a  house  with  three  tiers  of  windows, 
7  in  a  tier ;  the  height  of  the  first  tier  is  6  feet  11 
inches,  of  the  second  5  feet  4  inches,  and   of  the  third 

4  feet  3  inches  ;  the  breadth  of  each  window  is  3  feet 
6  inches  ;  what  will  the  glazing  come  to  at  14*d.  per 
foot  ?  Ms.  l.24<  8s.  5 Id. 

6.  What  will  the  glazing  a  triangular  sky -light 
come  to  at  lOd.  per  foot :  the  base  being  ±2  feet  6 
inches  long,  and  the  perpendicular  height  16  feet  9 
inches  ?  Ms.  Ia.  7s.  2^d. 

7.  What  is  the  area  of  an  elliptical  fan-light,  of  14 
feet 6  inches  in  length,  and  4  feet  9  inches  in  breadth? 

Ms.  68  feet  10  inches. 

8.  WThat  cost  the  covering  and  guttering  a  roof  with 
lead,  at  18s.  per  hundred-weight  ;  the  length  of  the 
roof  being  43  feet,  and  the  girt  over  it  32  feet ;  thfe 
guttering  being  57  feet  in  length  and  2  feet  in  breadth ; 
admitting  a  square  foot  of  lead  to  weigh  8f  pounds. 

Ms.  Z.104  15s.  3  |cL 


The  Mensuration  of  Masons-  Work;  §c. 


§VI.    Of  Ma  soy s'  Work.         • 

Masons  measure  their  work  sometime*  by  the  foot 
solid,  sometimes  by  the  foot  superficial,  and  sometimes 
by  ihe  foot  in  length.  In  taking  dimensions  they  girt 
all  their  mouldings  as  joiners  do. 

The  solids  consist  of  blocks  of  marble,  stone,  pil- 
lars, columns,  &c.  The  superficies  are  pavements, 
slabs,  chimney-pieces,  &c. 

Masons  reckon  all  such  stones  as  are  above  two 
inches  thick,  at  so  much  per  foot,  solid  measure.  And 
for  the  workmanship,  they  measure  the  superficies  of 
that  part  of  the  stone  which  appears  without  the 
wall. 

EXAMPLES. 

1.  If  a  wall  be  97  feet  5  inches  long,  18  feet  3 
inches  high,  and  2  feet  3  inches  thick;  how  many  solid 
feet  are  contained  in  that  wall  ? 


By  Duodecimals. 

F. 

I. 

97  : 

5  length. 

18  : 

3  breadth  or  height 

776 

97 

24  : 

4  :  3 

6  : 

0  :  o 

1  : 

6  :  0 

1777  : 

10  :  3  superficies. 

2  : 

3         thickness. 

3555  : 

8  :  6 

444  : 

5:6:9 

4000  :     2:0:9  solidity. 


By  Decimals. 

97.417 

18.25 

4S7085 
194834 
779336 
97417 

1777.86025 

2.25 

888930125 
355572050 
255572050 

4000.1855625 


The  Mensuration  of  Paviors'  Work.         245 

2.  If  a  wall  be  107  feet  9  inches  long,  and  20  feet 
6  inches  high ;  how  many  superficial  feet  are  contain- 
ed therein  ?  Jlns.  220$  feet  10  inches. 

3.  If  a  wall  he  112  feet  3  inches  long,  and  16  feet 
6  inches  high ;  how  many  superficial  rods  of  63  square 
feet  each,  are  contained  therein  ? 

Jlns.  29  rods  25  feet. 

4).  What  is  a  marble  slab  worth,  whose  length  is  5 
feet  7  inches,  and  breadth  1  foot  10  inches,  at  6s.  per 
foot  superficial  ?  Jlns.  1.3  Is.  5d. 


§  VII.  Of  Pa  viors*  Work. 

Paviors'  work  is  measured  by  the  square  yard,  con- 
sisting of  9  square  feet.  The  superficies  is  found  by 
multiplying  the  length  by  the  breadth. 

EXAMPLES. 

1.  What  cost  the  paving  a  foot  path  at  2s.  4d.  per 
yard ;  the  length  being  35  feet  4  inches,  and  breadth  S 
feet  3  inches  ? 


x  2 


246  Tfie  Mensuration  of  Paviors'  Work. 

By  Decimals.  Jhj  Duodecimals. 

F.         I. 
35.3383  &C  33      :      1 

8±  8:3 


382.6066  282     :     8 

8.8333  8     :  10 


9)291.4099  &C.  9)291 


3s.  id.is  1)32.3888  &C\  3s.  4d.  is  £)32.3    :     6 


J.5.3981  Feet     -     -     -     5    6     8 

20  3—-L  of  a  yard  0    1     1|33 
6=z^  of  3  feet    0    0     2.88 


S.  7.9620 


12  1.5    7  11'  21 


d.  11.5140 

4 


2.1760 

Answer  1.5  7s.  ll^d. 

2.  What  will  the  paving  a  court-yard  come  io  at 
3s.  id.  per  yard,  the  length  being  24  feet  5  inches, 
and  breadth  12  feet  7  inches?  dns.  1.5  13s.  9t]d. 

3.  What  will  be  the  expence  of  paving  a  rectangu- 
lar court-yard,  its  length  being  62  feet  7  inches,  and 
breadth  44  feet  5  inches ;  and  in  which  there  is  laid  a 
foot-path  the  whole  length  of  it.  54  feet  broad,  with 
flat  stones,  at  3s.  per  yard,  the  rest  being  paved  with 
pebbles  at  2s..  6d.  per  yard?  dns.  I.B9  lis.  S^d1. 


Of  the  Sliding  Rule,  2if 


CHAPTER  V. 


Of  Gauging*. 


GAUGING  is  the  art  of  measuring  and  finding  tlie 
contents  of  all  sorts  of  vessels,  in  gallons,  or  cubi£ 
inches ;  such  as  casks,  brewers'  vessels,  &c.  &c» 


Of  the  Sliding  Rule. 

The  sliding  rule  is  an  instrument,  particularly  use- 
ful in  gauging,  made  generally  of  box,  in  the  form  of 
a  parallelopipedon.  There  are  various  kinds,  but  the 
the  most  convenient,  or  at  least  that  which  is  most 
used  in  the  excise,  was  invented  by  Mr.  Verie,  collec- 
tor of  the  excise. 

1st,  The  line,  marked  A,  on  the  face  of  this  rule, 
is  called  Gunter's  Ride,  and  is  numbered  1.  2.  3.  4.  5. 
6.  7.  8.  9.  10.  At  2150.42  is  a  brass  pin,  marked  MB, 
signifying  the  cubic  inches  in  a  bushel  of  malt ;  at  282 
is  another  brass  pin,  marked  A,  signifying  the  number 
of  cubie  inches  in  a  gallon  of  ale. 

2dly,  The  line  marked  B  is  on  the  slide,  and  is  di* 
vided  in  exactly  the  same  manner  as  that  marked  A ; 
there  is  another  slide  B,  which  is  used  along  with 
this ;  the  two  brass  ends  are  then  placed  together,  and 
so  make  a  double  radius  numbered  from  the  left-hand 
towards  the  right.    At  231,  on  the  second  radius,  is  & 


'„>is  Of  the  Sliding  Rule. 

brass  pin,  marked  AV,  signifying  the  cubic  inches  in  a 
gallon  of  wine;  at  314-  is  another  brass  pin,  marked 
C,  signifying  the  circumference  of  a  circle  whose 
diameter  is  1.  The  manner  of  reading,  and  using 
these  lines,  is  exactly  the  same  as  the  lines  A  and  13, 
described  in  chap.  XI.  page  52,  &c. 

3dly,  The  back  of  the  first  radius,  or  slide,  marked  B, 
contains  the  divisors  for  ale,  wine,  mash-tun  gallons, 
malt,  green  starch,  dry  starch,  hard  soap  hot,  hard  soap 
cold,  green  soft  soap, white  soft  soap,  flint  glass,  Sfc.  Sfc. 
as  in  the  table,  prob.  1,  following. — Tlie  back  of  the 
second  radius,  or  slide,  marked  B,  contains  the  gauge- 
points  correspondent  to  these  divisors,  where  S  stands 
for  squares,  and  C  for  circles. 

^thly,  The  line  M.  D.  on  the  rule,  signifying  malt 
depth,  is  a  line  of  numbers  beginning  at  2150.42,  and 
is  numbered  from  the  left  to  the  right-hand  2.  10.  9.  8. 
7.  6.  5.  4.  3.     This  line  is  used  in  malt  gauging. 

5thly,  The  two  slides,  B,  just  describ  d,  are  always 
used  together  either  with  the  line  A;  Ml);  or  the 
line  D,  which  is  on  the  opposite  face  of  the  rule  to 
that  already  described.  This  line  is  numbered  from 
the  left-hand  towards  the  right  1.  2.  3.  31.  32.  which 
is  at  the  right-hand  end:  it  is  then  continued  from 
the  left-hand  end  of  the  other  edge  of  the  rule  32.  4. 
B.  6.  7.  8.  9.  10.  At  17.15  is  a  brass  pin  marked  WG. 
signifying  the  circular  gauge-point  for  wine  gallons. 
At  18.95  is  a  brass  pin  marked  AG  for  ale-gallons. 
At  46.37  MS  signifies  the  square  gauge  point  for 
malt  bushels.  At  52.32  MR  signifies  the  round,  or 
circular  gauge-point  for  malt  bushels.  The  line  D  on 
this  rule  is  of  the  same  nature  as  the  line  marked  D 
on  the  Carpenter's  Bule,  described  in  chap.  XL  page 
52.  The  line  A  and  the  two  slides  B  are  used  toge- 
ther, for  performing  multiplication,  division,  propor- 
tion. &c.  and  the  line  D,  and  the  same  slides  B,  are 
used  together  for  extracting  the  square  root,  &c. 


Of  the  Sliding  Rule.  249 

Uhly,  The  other  two  slides  belonging  to  this  rule 
are  marked  C,  and  are  divided  in  the  same  manner, 
and  used  together,  like  the  slides  B.  The  back  of  the 
first  radius  or  slide,  marked  C,  is  divided,  next  the 
edge,  into  inches,  and  numbered  from  the  left-hand 
towards  the  right  1.  2.  3.  4.  &c.  and  these  inches  are 
again  divided  into  ten  equal  parts.  The  second 
line  is  marked  spheroid,  and  is  numbered  from  the 
left-hand  towards  the  right  1.  2.  3.  4.  5.  6.  7.  The 
third  line  is  marked  second  variety,  and  is  numbered 
1.  2.  3.  4.  5.  6.  These  lines  are  used,  with  the  scale 
of  inches,  for  finding  a  mean  diameter,  see  prob.  X. 
following. 

The  third  and  fourth  variety  are  omitted  on  this 
slide,  and  with  good  reason,  for  it  is  very  probable 
that  there  never  was  a  cask  made  resembling  either 
of  these  forms.  The  back  of  the  second  radius,  or  slide, 
marked  C,  contains  several  factors  for  reducing  goods 
of  one  denomination  to  their  equivalent  value  in  those 
of  another.  Thus  |  X  to  VI  6.  |  signifies  that  to  re- 
duce strong  beer  at  8s.  per  barrel  to  small  beer  at 
Is.  4d.  you  must  multiply  by  6.  |  VI.  to  X.  17.  |  signi- 
fies that  to  reduce  small  beer  at  Is.  4d.  per  barrel 
to  strong  beer  at  8s.  per  barrel,  you  must  multiply 
by  .17.  |  C  4s.  to  X.  .27.  |  signifies  that  27  is  the  mul- 
tiplier for  reducing  cider  at  4s.  per  barrel  to  strong 
beer  at  8s.  &c. 

7thly,  The  two  slides  C,  just  described,  are  always 
used  together,  with  the  lines  on  the  rule  marked  seg. 
st.  or  88,  segments  standing ;  and  seg.  ly  or  SL,  seg- 
ments, lying,  for  ullaging  casks.  The  former  of  these 
lines  is  numbered  1.  2.  3.  4.  5.  6.  7.  8,  which  stands 
at  the  right-hand  end ;  it  then  goes  on  from  the  left- 
hand  on  the  other  edge  S.  9.  10,  &c.  to  100 ;  the  latter 
is  numbered  in  a  similar  manner,  1.  2.  3.  4,  which 
stands  at  the  right-hand  end ;  it  then  goes  on  from  the 
left-hand  on  the  other  edge  4.  5.  6.  to  100,  &c 


23%  Of  Gauging. 


PROBLEM    I. 


To  find  the  several  Multipliers,  Divisors,  and   Gauge- 
points,  belonging  to  the  several  Measures  now  us 
England. 


\-v 


For  Square  figures,  the  following  multipliers  and 
divisors  are  to  be  used: 

1.0000(.0Q3546  multiplier  for  ale  gallons. 

231)1.0000(.00l: .29  multiplier  for  wine  gallons. 

268. S)i.OOO(. 0037 202  multiplier  for  malt  gallons. 

2150.42)1.00()(.0004f)502  multiplier  for  malt  bushels. 

227)1. 000(. 00405  multiplier  for  mash-tun  gallons. 

So,  if  the  solid  inches  in  any  vessel  be  multiplied  by 
the  said  multipliers,  the  product  will  be  the  gallons  in 
the  respective  measures  ;  or  dividing  by  the  divisors 
2S2,  231,  or  268.3  ;  the  quotient  will  likewise  be  gal- 
lons. 

Note.  That  282  solid  inches  is  a  gallon  of  ale  or 
beer  measure :  231  solid  inches  is  a  gallon  of  wine 
measure  ;  268.8  solid  inches  is  a  gallon,  and  2150.42 
solid  inches  is  a  bushel  of  malt,  or  corn  measure. 

For  circular  areas,  the  following  multipliers  and  di- 
visors are  to  be  used : 

282).785398(. 002785  multiplier  for  ale  gallons.        + 

231).785398(. 003399  multiplier  for  wine  gallons,     -f- 

.785398)282.(359.05  divisor  for  ale  gallons. 

.785398)281.(294.12  divisor  for  Mine  gallons. 

.785398)2150.42(2738  divisor  for  corn  bushels. 

The  square  root  of  the  divisor  is  the  gauge-point. 

r^x  .   .    fAle  measure,  is  16.79 

The    gauge-point  J  Wine  mea8u;     is  ±5±Q 

for  squares  in     j Ma,t  ^^  .'  ^7 

The  gauge-point    fAle  measure,  is  18.95 

for  circular  fi-  -<  Wine  measure,  is  17.15 

gures  in  ^Malt  bushel,  is  52.32 

And  thus  the  numfljBin  the  following  table  were  cal- 
culated. 

IT** 


Of  Gauging 


>31 


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Note.  It  often  happens  in  the  practice  of  gauging, 
that  when  the  one  given  number  is  set  to  the  gauge- 
point  on  the  sliding  rule,  the  other  given  number  will 
fall  oft*  the  rule ;  hence  in  many  cases  it  will  be 
necessary  to  find  a  second,  or  new  gauge-point. — 
The  second  gauge-points  are  the  square  roots  of  10 
times  the  divisors  in  the  above  table.  Thus  for  squares, 


Of  Gauging. 

the  new  gauge-point  for  ale  is  53.10,  for  wine  48.06j 
for  malt-misheU  L 4.66  5  ami  for  circles,  the  new  gauge- 
point  for  ale  is  02.92,  for  wine  94.22,  and  lor  malt- 
bushels  ig.54. 

By  the  Sliding  Rule. 

Set  1  on  B,  to  the  old  gauge-point  on  D,  and  against 
the  other  1  on  B,  is  the  new  gauge-point  on  1). 


l'HOBLEM    II. 

To  find  the  Area  in  Me  or  Wine  Gallons,  of  any  rectilin- 
eal plane  Figure,  whether  triangular,  quadrangular, 
or  midtangular. 

To  resolve  this  problem,  you  must,  by  chap.  I.  part 
II.  find  the  area  in  inches,  and  then  bring  it  to  gallons, 
by  dividing  that  area  in  inches  by  the  proper  divisor, 
viz.  by  282  for  ale,  or  by  231*for  wine  ;  or  else  by  mul- 
tiplying by  .003546  for  ale,  or  by  .001329  for  wine  ;  and ' 
the  quotient  or  product  will  be  the  area. 

EXAMPLES. 

1.  Suppose  a  back  or  cooler  in  the  form  of  a  paral- 
lelogram, 250  inches  in  length,  and  84.5  inches  in 
"breadth  ;  what  is  the  area  in  ale  or  wine  gallons  ? 

Multiply  250  by  84.5,  and  the  product  is  21125, 
the  area  in  inches,  which  divide  by  282,  and  the  quo- 
tient is  74.9  gallons  of  ale  ;  or  multiplied  by  .003546, 
the  product  is  74.90925  gallons,  nearly  the  same  5  and 
if  2112.5  be  divided  by  231,  or  multiplied  by  .004329, 
it  will  give  91.45  gallons  of  wine. 

Note.  The  areas  of  all  plane  figures,  in  gauging, 
arc  expressed  in  gallons  ;  because  there  will  be  the 


Of  Gauging.  25 


3& 


same  number  of  solid  inches  in  any  vessel  of  one  inch** 
deep,  as  there  are  superficial  inches  in  its  base ;  so  that's^ 
what  is  called  by  gaugers  a  surface,  or  area,  is  in  re-  °* 
ality  a  solid  of  one  inch  in  depth.  N^t 

By  the  Sliding  Rule. 

on  A.  on  B.        on  A.      on  B. 

As  282     :     84.5     ::    '250     :     74.9 
As  231     :     S4.5     ::     250     :     91.45 

2.  If  the  side  of  a  square  be  40  inches,  what  is  the 
area  in  ale  gallons  ?  Jlns.  5.67  gallons. 

3.  The  longest  side  of  a  parallelogram  is  50  inches; 
the  shortest  side  30  inches;  what  is  the  area,  in  ale 
gollons  ?  Jlns.  5.32  gallons. 

4.  The  side  of  a  rhombus  is  40  inches,  and  its  per- 
pendicular breadth  37  inches ;  what  is  its  area  in  ale 
gallons  ?  Jlns.  5.25  gallons. 

5.  The  length  of  a  rhomboides  is  48  inches,  and  its 
perpendicular  breadth  32  inches  ;  what  is  its  area  in 
wine  gallons  ?  Jlns.  6.65  gallons.    . 

6.  The  base  of  a  triangle  is  60  inches,  and  its  per- 
pendicular 23.5  inches ;  required  its  area  in  ale  gal- 
lons ?  Jlns.  2.5  gallons. 

7.  The  diagonal  of  a  trapezium  is  60  inches,  and 
the  two  perpendiculars  from  the  opposite  angles  15  and 
27  inches  ;  what  is  its  area  in  ale  gallons  ? 

Jlns.  4.47  gallons. 

8.  The  side  of  a  pentagon  is  50  inches,  what  is  its 
area  m  ale  gallons  ?  Jlns.  ±5.25  gallons. 

9.  The  side  of  a  hexagon  is  64  inches,  and  the  per- 
pendicular from  the  centre  to  the  middle  of  one  of  the 
sides  55.42 inches  ;  required  its  area  in  ale,  wine  gal- 
Ions,  and  malt  bushels  ? 

.73  ale. gallons. 


f37. 

;.  <  46. 
I   4. 


tins.  4  46.06  wine  gallons. 
94  malt  bushels. 
Y 


(if  Gauging. 


PROBLEM    III. 

The  Diameter  of  a  Circle  being  given  in  inches,  to  find 
the  Area  of  it  in  Ale  or  Wine   Gallons,  <S(c. 

If  the  square  of  the  diameter  be  multiplied  by 
.00278a  for  ale,  or  by  .003399  for  wine;  or  if  it  be  di- 
vided by  359.05  for  ale,  or  by  294.12  for  wine,  the  pro- 
ducts or  quotients  will  be  the  respective  ale  or  wine 
gallons  :  for  any  other  denomination,  use  the  proper 
multiplier  or  divisor  in  the  table. 

EXAMPLES. 

1.  Suppose  the  diameter  of  the  circle  be  32.6  inches; 
what  will  be  the  area  in  ale  and  wine  gallons  ? 

The  square  of  32.6  is  1062.76. 
Then  339.05)1062.76(2.2592  area  in  ale  gallons. 
And  294.12)1062.76(3.6133  area  in  wine  gallons. 
Or  16152.76 X  .002785— 2.9398  ale  gallons. 
And  1062.76X- 0034=3. 6133  wine  gallons. 

By  the  Sliding  Rule. 

on  D.     on  B.         on  D.       on  B. 
As  18.95     :     1     :  :     32.6     :     2.96 
As  17.13     :     1     :  :     32.6     :     3.61 
The  two  first  terms  are  the  circular  gauge-points; 
see  the  table. 

2.  If  the  diameter  of  a  circle  be  45  inches,  what  is 
its  area  in  ale  gallons  ?  dns.  5.64  gallons. 

3.  If  the  diameter  of  a  circle  be  68  inches;  required 
its  area  in  ale  and  wine  gallons,  and  malt  bushels  ? 

.2.87  ale  gallons. 


f  12.8 
;.  -j  15.7 
I    1.6i 


Jins.  ■{  15J72  wine  gallons. 
68  malt  bushels. 


Of  Gauging 


TROBLEM    IV 


The  Transverse,  or  Longer  Diameter,  and  the  Conju- 
gate, or  Shorter  Diameter,  of  an  Ellipsis  being  given , 
to  find  its  Area  in  Ale  or  Wine  gallons. 

If  the  rectangle  or  product  of  the  two  diameters, 
that  is,  of  the  length  and  breadth  of  an  ellipsis,  be  di- 
vided by  359.05,  or  multiplied  by  .002785  for  ale,  or 
divided  by  294.12,  or  multiplied  by  .0034  for  wine,  the 
quotient  or  product  will  be  the  ale  or  wine  gallons  re- 
quired. And  for  any  other  denomination  use  the  pro- 
per divisor  or  multiplier. 


EXAMPLES. 

1.  Suppose  the  longer  diameter  be  81.4  inches,  and 
the  shorter  diameter  be  54.6  inches;  what  will  be  the 
area  of  that  ellipsis  in  ale  and  wine  gallons  ? 

Multiply  81.4  by  54.6,  and  the  product  is  4444.44 ; 
then, 

359.05)4444.44(12.38  area  in  ale  gallons. 
294.12)4444.44(15.11  area  in  wine  gallons.. 
Or  4444.44 X.002785zzl2.38  ale  gallons. 
And  4444,44X0034—15.11  wine  gallons. 

By  the  Sliding  Rule. 


on  A. 

on  B. 

on  A 

onB. 

As  359 

:     81.4     : 

:     54.6 

:     12.4  ale  gallons. 

As  294 

:     81.4     : 

:     54.6 

:     15.1  wine  gallons 

2.  The  transverse  diameter  of  an  ellipsis  is  72  inch- 
es, conjugate  50  inches ;  what  is  its  area  in  ale  gallons  ? 

dns,  10  gallons. 


'-'.m'»  Of  Gauging. 

3.  If  the  transverse  diameter  of  an  ellipsis  be  70 
inches,  conjugate  50  inches ;  what  is  its  area  in  ale 
and  wine  gallons,  and  malt  bushels  ? 

f   9.7 1<  ale  gallons. 
,1ns.  <  It. 90  wine  gallons. 
(    1.27  malt  bushels. 


PROBLEM    V. 

To  find  the  Content  in  Me  or  Wine    Gallons  of  any 
Prism,  whatsoever  form  its  base  is  of. 

First,  find  its  solid  content  in  inches  (by  sect.  I. 
II.  III.  of  chap.  II.  part  II.)  then  divide  that  content 
in  inches  by  2S2  for  ale,  or  by  231  for  wine;  the 
respective  quotients  will  be  the  content  in  ale  or  wine 
gallons. 

Otherwise,  you  may  find  the  content  of  a  prism,  by 
finding  the  area  of  its  base  in  gallons  (by  problem  II 
of  this  chapter)  and  multiply  that  area  by  the  vessel's 
height,  or  depth  within,  the  product  will  be  its  eontent 
in  gallons. 

EXAMPLES. 

1.  Suppose  a  vessel,  whose  base  is  a  right-angled 
parallelogram,  its  length  being  49.3  inches,  its 
breadth  36.3  inches,  and  the  depth  of  the  vessel  is 
42.6  inches;  the  content  in  ale  and  wine  gallons  is  re- 
quired ? 

The  length,  breadth,  and  depth,  being  multiplied 
continually,  the  product  is  76656.57;  which  divided  by 
282,  the  quotient  is  271.83  ale  gallons  :  and  divided 
by  231,  the  quotient  is  331.84  wine  gallons:  and  by 
dividing  by  2152.42,  such  a  cistern  will  be  found  to 
hold  35.65  bushels  of  corn. 


Of  Gauging,  257 

By  the  Sliding  Rule. 

on  B.  on  D.  on  B.  on  D. 

As  the  length  :  length  : :  breadth  :  mean  proportion 

on  D.        on  B.  on  D.  on  B. 

As  sq.  gauge  p.  :  depth  :  :  mean  proportion  :  content 
on  B.       on  D.        on  B.         on  D. 

viz.  49.3     :     49.3     :  :    36.5     :     42.42 
on  D.  on  B. 

16.: 

15.19 

46.37  \    '  rc""J    '  '    ^*^  *  /     35.65  malt  bush. 


1      onB.        on  D.  f™  al.e  SaI-, 

\    :43.6    =:    42.42  08f-*WTfl 
J  (^    35.65  malt  bus! 


2.  Eaeh  side  of  the  square  base  of  a  vessel  is  40 
inches,  and  its  depth  10  inches ;  what  is  the  content 
in  ale  gallons  ?  Jins.  56.7  gallons. 

3.  The  length  of  a  rectangular  parallelopipedon  is 
72  inches,  breadth  33  inches,  and  depth  82  inches; 
required  the  content  in  ale  and  wine  gallons,  and  malt 
bushels  ?  ("690.89  ale  gallons. 

*flns.  <  843.42  wine  gallons. 
(_   90.61  malt  bushels. 

4.  The  diameter  of  a  cylindrical  vessel  is  32  inches, 
the  internal  depth  45.5  inches ;  required  its  content  in 
ale  and  wine  gallons,  and  malt  bushels  ? 

129.78  ale  gallons. 
Jlns.  ^  158.47  wine  gallons. 
17.01  malt  bushels. 


{ 


PROBLEM  VI. 


To  find  the  Content  of  any  Vessel  whose  ends  are 
Squares,  or  Rectangles  of  any  dimensions. 


RULE. 


Multiply  the  sum  of  the  lengths  of  the  two  ends  by 
the  sum  of  their  breadths,  to  which  add  the  areas  of 
the  two  ends  ;  this  sum  multiplied  by  one-sixth  of  the 

Y   2 


408  Of  Gauging. 

depth  will  give  the  solidity  in  cubic  inches;  which 
divided  hj  282.281,  or  2100.42  will  give  the  content 
in  ale  gallons,  wine  gallons,  or  malt  bushels. 

EXAMPLES. 

1.  Suppose  a  vessel,  whose  bases  are  parallelograms : 
the  length  of  the  greater  is  100  inches  and  its  breadth 
TO  inches  ;  the  length  of  the  less  base  80,  and  its 
breadth  56,  and  the  depth  of  the  vessel  42  inches; 
the  content  in  ale  and  wine  gallons  is  required? 

180=100-|-80,  sum  of  the  lengths    ~)    of  the  two 
126=  70-J256,  sum  of  the  breadths  5        ends. 

32680  product. 
7000  area  of  the  greater  base— 100X70. 
4480  area  of  the  less  base:=S0X56. 


34160  sum. 

7  one-sixth  of  the  depth. 


239120  solidity  in  cubic  inches. 

282)239120(847.94  ale  gallons. 
'J.Z  1  )239120(1035.15  wine  gallons. 

By  the  Sliding  Rule. 

Find  a  mean  proportional  (83.66)  between  the  lengfh 
and  breadth  at  the  greater  end,  and  a  mean  propor- 
tional (66.93)  between  the  length  and  breadth  at  the 
less  end ;  the  sum  of  these  is  150.59,  twice  a  mean 
proportional  between  the  length  and  breadth  in  the 
middle.     Then 

.  on  D.        on  B.         on  D.  on  B. 

16.79  :  V  :  :  83-66  :  173-7 
:  :  66.93  :  111.2 
:  :  150.59     :     563. 


847.9  A.  G. 


Of  Gauging.  259 

The  content  in  wine  gallons  may  be  found  by  using 
the  wine  gauge-point. 

2.  Each  side  of  the  bottom  of  a  vessel,  in  the  form 
of  the  frustum  of  a  square  pyramid,  is  27  inches,  each 
side  at  the  top  13.8  inches,  and  the  depth  2\  inches; 
required  the  content  in  ale  gallons  ? 

Jins.  32.07  gallons, 
• 

3.  There  is  a  tun,  whose  parallel  ends  are  rectan- 
gles, the  length  and  breadth  of  the  top  36  and  32 
inches ;  the  length  and  breadth  at  the  bottom  48  and 
40  inches,  and  the  depth  60  inches :  required  the  con- 
tent of  the  tun  in  ale  gallons,  wine  gallons,  and  malt 
bushels  ?  f  323.41  ale  gallons. 

Jlns.  <  394.80  wine  gallons. 
i    42.4    malt  bushels-. 


PROBLEM    VII. 


To  find  the  Content  of  a  Vessel,  whose  bases  are  paral- 
lel and  circular,  it  being  the  Frustum  of  a  Cone. 


RULE. 


To  three  limes  the  product  of  the  two  diameters,, 
add  the  square  of  their  difference,  multiply  the  sum 
by  one-third  of  the  depth,  and  divide  the  product  by 
the  proper  circular  divisor,  viz.  359.05  for  ale,  291.12 
for  wine,  and  2738  for  malt  bushels,  &c.  and  the  quo- 
tient will  be  tjie  content  accordingly. 


EXAMPLES. 

1.  Suppose  ihe  greater  diameter  80  inches,  and  the 
less  diameter  71  inches,  and  the  depth  34  inches,  the 
content  in  ale  and  wine  gallons  is  required  ? 


SCO  Of  Gauging. 

80  greater  diameter.  71  less  diameter. 

71  less  diameter.  80  greater  diam. 


9  difference.  5680  product. 

9  3 


St  square  of  the  diff.  17040 

—  81  square  of  diff. 

3)  171 21  sum. 

359.05)194031(540.42  A.  G. 

294.12)19403S(659.72  W.  G.     5707 

34  depth. 

194038  product. 

In  the  above  example,  instead  of  multiplying  the 
sum  by  one-third  of  the  depth,  one-third  of  the  sum 
is  multiplied  by  the  whole  depth,  which  amouuts  to 
the  same. 

By  the  Sliding  Rule. 

Find  a  mean  proportional  between  the  two  diame- 
ters: Thus, 

on  B.     onD.        on  B.       onD. 
80     :     80     :  :     71     :    75.36;  again 
Set  18.95  on  D  to  il|  (one-third  of  the  depth)  on  B. 
And  against  71  on  D  stands  the  content  on  B  159.09 
against  75.36  on  D  stands  the  content  on  B  179.22 
»  stands  the  content  on  B  202.00 

Sum  is  the  whole  content  540.31 


against  80       on  D  stands  the  content  on  B  202.00 


In  the  same  manner  the  content  in  wine  gallons  may 
be  fouud,  using  the  wine  gauge-point. 

2.  The  greater  diameter  of  a  conical  frustum  is  38 
inches,  the  less  diameter  20.2,  and  depth  .21  inches  ; 
required  the  content  in  ale  gallons  ? 

Ans.  51.07  gallons. 


I 


Of  Gauging.  261 


The  top  diameter  of  a  conical  frustum  is  22 
inches,  bottom  diameter  40  inches,  and  the  depth  60 
inches ;  required  the  content  in  ale  gallons,  wine  gal- 
lons, and  malt  bushels  ?  f  163.12  ale  gallons. 


f  163.12 
>.  -j  201.55 
(_    21.63 


dns.  -(  201.55  wine  gallons, 
malt  bushels. 


PROBLEM   VIII. 


To  Gauge  and  Inch  a  Tun  in  the  form  of  the  Frustum 
of  a  Cone,  and  to  make  an  allowance  for  the  drip  or 
fall. 

The  inching  of  a  tun,  or  vessel,  is  finding  how  much 
liquor  it  will  hold  at  every  inch  of  its  depth. 

When  a  vessel  does  not  stand  even,  or  with  its 
base  parallel  to  the  horizon,  the  quantity  of  liquor 
mDw,  which  will  just  cover  the  bottom,  is  called  the 
drip  or  fall  of  the  tun. 

Let  water  be  poured  into  a  tun  till  the  bottom  is 
just  covered,  which  will  be  when  the  water  touches 
the  point  m,  and  suppose  the  measure  of  the  water  in 
this  case  to  be  30.92  gallons. 

Find  the  horizontal  line  FB,  which  will  be  parallel 
to  m  n9  the  surface  of  the  liquor ;  the  line  FB  will 
represent  the  surface  of  the  liquor  when  the  vessel  is 
full. 

Find  the  perpendicular  depth  on  to  the  surface  of 
the  liquor,  which  suppose  to  be  26  inches. 

Take  mean  diameters  pa-     .A^— " — ' — 

rallel  to  FB  and  m  n  at  every       Ffc^  "%""      fv^Nr 

6  or  10  inches  depth.     Sup-        tf,fi§^§:- mSSff 

pose   the  first  mean   diame-  filliP  SMlf  * 

ter  a  a  at  5  inches  from   FB        ifillL--       ^8l~h 
to  be  83.6  inches;  the  second  lERi       "jjl|l>f 

mean  diameter  b  b  at  13  inch-      _g6l|g^^;ejfe^ 
es  from  FB—7S.7  inches  and  *"eg^Sii||||^ 

the  third  mean  diameter  c  c 
at  23  inches  from  FB=74.3  inches. 


8M  Of  Gait" 

Find  the  areas  correspondent  to  each  of  those  di- 
ameters, as  in  the  third  column  of  the?  following  ta- 
ble; multiply  these  areas  by  their  respective  depths 
and  you  will  have  the  fourth  column.  Lastly,  these 
contents  being  brought  into  barrels,  &e.  allowing  24 
gallons  to  a  barrel,  and  8*-  gallons  to  a  firkin,  will  give 
the  remaining  columns. 


Depths. 

Diam. 

Area. 

Content. 

Content. 

in  gallons. 

B. 

F. 

G. 

oP=lo 
PS=lo 
Sw=  6 

S3.6 

78.7 
74.5 

19.465 
17.250 
15.458 

194.65 

172.50 

92.75 

30.92 

5 

2 
0 

0 

2 

3 

7.65 
2.50 

7.75 

5.42 

whole  26 

Cont.of  the  drip. 

Whole  content. 

490.82 

:  i 

1  ft. 32 

Now  to  find  the  content  at  every  inch  of  the  depth, 
reduce  the  first  area  19.465  into  barrels,  &c.  and  it 
will  be  2  F.  2.465  G.  which  subtracted  from  the 
whole  content  14  B.  1  F.  6.32  G.  leaves  13  B.  3  F. 
3.8555  G.  the  content  when  1  inch  is  dry;  and  thus 
continue  subtracting  the  first  area  from  the  several 
remainders  until  10  inches  are  dry,  when  there  will 
remain  8  B.  2  F.  7.17  G. — Then  take  the  second 
area  17.25,  having  first  brought  it  into  firkins,  &c. 
and  subtract  it  from  the  last  remainder,  and  thus 
continue  to  do  till  you  have  20  inches  dry,  when  there 
will  remain  3  B.  2  F.  4.67  G — Lastly,  take  the  third 
area  15.458,  having  first  brought  it  into  firkins,  &c. 
and  subtract  it  from  the  area  when  20  inches  are 
dry ;  and  thus  proceed  till  you  have  26  inches  dry, 
and  there  will  remain,  if  you  have  made  no  mistake, 
3  F.  5.42  G.  the  quantity  of  liquor  contained  in  the 
drip  or  fall. 


Of  Gauging. 


263 


The  point  F  may  be  marked  as  a  constant  dipping 
place,  and  the  several  contents  entered  into  a  dimen- 
sion book.  Then  to  find  how  much  liquor  there  is 
in  this  fixed  vessel,  at  any  future  time,  take  the  depth 
of  the  liquor,  and  against  that  depth  in  your  table  you 
will  find  the  content. 


PROBLEM    IX. 

To  gauge  a  Copper. 

Let  ABCD  be  a  small  copper  to  be  gauged. 

Take  a  small  cord  of  packthread,  make  one  end 
fast  at  A,  and  extend  the  other  to  the  opposite  side 
of  the  copper  at  B,  where  make  it  fast.  Then  with 
some  convenient  instrument  take  the  nearest  distance 
from  the  deepest  place  in;the  copper,  to  the  thread,  as 
«C,  which  suppose  to  be  47  inches. 


In  like  manner,  set  the  end  of  the  instrument  or  rule 
upon  the  top  of  the  crown  at  d.  and  take  the  nearest 
distance  to  the  thread,  as  dg,  which  suppose  4.2  inches  : 
this  subtracted  from  aC,  47,  the  remainder  5  is  the 
altitude  of  tUf  crown. 


j it  l  Of  Gaugi 

To  find  CD,  the  diameter  of  the  hoi  loin  of  the 
crown. 

Measure  AB,  the  diameter  ef  the  top,  which  admit 
to  hi-  y(.>  Inches;  then  hold  a  thread  so  as  a  plummet 
at  the  end  thereof  may  hang  just  over  C,  by  which 
means  you  will  find  the  distance  Aa.  Do  the  like  on 
the  other  side:  so  will  yOu  find  also  the  distance, 
which  suppose  17.5  inches  each;  add  these  two  togeth- 
er, and  subtract  their  sum  (viz.  35)  from  99,  and  the 
remainder  is  04  inches,  equal  to  CD,  the  diameter  at 
the  bottom 'of  the  crown.  The  diameter  hk9  which 
touches  the  top  of  the  crown,  may  be  found,  by  mea- 
suring, to  be  65  inches. 

Now  to  find  the  content  of  the  copper  from  tlie 
crown  upwards,  that  is  the  part  ATM//,  the  depth  i  ' 
being  42  inches,  you  may  take  the  diameter  in  the  mid 
die  of  every  6  inches  of  the  depth,  which  suppose  to 
he  as  in  the  second  column  of  the  following  table,  the 
numbers  in  the  third  column  are  the  respective  areas 
in  ale  gallons,  found  by  prob.  ID.  the  fourth  column 
shews  the  content  of  every  6  inches  ;  all  which  being 
added  together,  the  sum  will  be  the  content  of  that 
part,  AB/i7i ;  that  is,  so  much  as  it  will  hold  after  the 
crown  is  covered. 

Now,  if  the  crown  be  taken  for  the  segment  of  a 
sphere, the  content  (by  the  latter  part  of  sect.  XI.  p. 
74.)  will  be  found  to  be  28.75  gallons. 

But  may  be  more  readily  found,  very  near  the  truth, 
thus: 

The  diameter  CD  was  found  to  be  64,  and  the  area 
to  this  diameter  is  11.408;  this  multiplied  by  half  the 
crown's  altitude,  viz.  by  2.5,  gives  28.52  gallons,  the 
content  of  the  crown. 

The  content  of  the  part  hkiDC  is  57.935  gallons ; 
from  which  subtract  the  content  of  the  crown,  28.52, 
and  the  remainder  is  29.415  gallons,  and  so  much  li- 
quor will  just  cover  the  crown. 


Of 

Gauging. 

265 

Parts 
of  the 
depth. 

Diam. 

Areas. 

Content  of 

every  six 

inches. 

6 
6 
6 
6 
6 
6 
6 

i>5.3 
90.1 
85.0 
80. 

75.2 
70.5 
66. 

25.2943 
22.6095 
20.1223 
17.8246 
15.7499 
13.8426 
12.1310 

151.767 

135.657 

120.734 

106.947 

94.499 

8   .056 

72.791 

Thesu 
To  jus 

765  4  1 

t  cover  the  crown     -     - 
hole  content     -     -     -     - 

29.415 

Thew 

-     794.866 

The  contents  in  the  last  column  may  be  brought  in- 
to firkins  and  barrels,  and  then  the  content  at  every 
inch  in  depth,  as  in  the  Vlllth  problem,  may  be  found 
and  entered  into  the  dimension  book. 


PROBLEM    X. 


To  compute  the  Content  of  any  close  Cask, 

In  order  to  perform  this  difficult   part   of  gauging, 
te  three  fol 
truly  taken  : 


the  three  following  dimensions  of  the  cask  must  be 


!    } 

cask,  J 


within  the  cask. 


f  The  bung-diameter, 
Viz.  <  The  head-diameter 
(The  length  of  the 

In  taking  these  dimensions,  it  must  be  carefully'ob- 
served, 

1st,  That  the  bung-hole  be  in  the  middle  of  the 
cask ;  also,  that  the  bung-stave,   and  the  stave  op- 


Of  Gauging, 

posite  to  the  bung-hole,   arc   both  regular  and 
within. 

2dljfj  That  the  heads  of  the  en-k'-  arc  equal,  and 
truly  circular:  if  so.  the  distance  between  the  inside 
of  the  chhnb  to  the  outside  of  the  opposite  stave  will 
be  the  head-diameter  within  the  cask,  ven  near. 

The  diameters  and  length  of  one  cask  may  be  equal 
to  those  of  another,  and  yet  one  of  those  casks  may 
contain  several  gallons  more  than  the  other. 

\>  for  instance, the  figure 
MH'DF  is  supposed  to  rep- 
resent a  cask  :  then  it  is 
plain,  that  if  the  outward 
curve  lines,  ABC,  and  FGD, 
are  the  hounds  or  staves  of 
the  cask,  it  will  hold  more 
than  if  the  inner  dotted  lines  """  '&' 

were  the  hounds,  or  staves ;  and  yet  the  bung-diame- 
ter BG,  and  head-diameters  CD  and  AF,  and  the 
length  LH,  are  the  same  in  both  those  casks. 

Whence  it  appears,  that  no  one  general  rule  can  be 
given,  by  which  all  sorts  of  casks  may  be  gauged ;  and 
therefore  gaug«>rs  usually  suppose  every  cask  to  be  in 
some  of  these  forms  : 

ls£,  The  middle  zone  or  frustum  of  a  spheroid  ;  see 
page  190. 

2dly,  The  middle  frustum  of  a  parabolic  spindle  ; 
see  page  18. 

Sdly,  The  lower  frustums    of  two  equal  parabolic 
conoids ;  see  page  193. 
(  4<thly,  The  lower  frustums  of  two  equal  cones,  such 
as  the  figure  represented  by  the  dotted  lines  above. 

5thly,  The  contents  of  casks  are  sometimes  found  by 
having  a  mean  diameter  between  the  bung  and  head 
given ;  see  the  rule  at  page  200. 


Of  Ganging.  267 

find  the  Content  of  a  Cask  by  the  mean  Diameter. 


RULE. 

Multiply  the  difference  between  the  head  and  bung- 
diameters,  when  it  is  less  than  6  inches,  by  .68  for  the 
first  variety  5  by  .62  for  the  second  variety  ;  by  .55  for 
the  third;  and  by  .5  for  the  fourth.  Or,  when  the  dif- 
ference between  the  head  and  bung-diameters  exceeds 
6  inches,  multiply  that  difference  by  .7  for  the  first  va- 
riety ;  .64  for  the  second  ;  .57  for  the  third  ;  and  .52 
for  the  fourth.  Add  this  product  to  the  head  diame- 
ter, and  that  sum  will  be  a  mean  diameter.  Square  the 
mean  diameter,  and  multiply  that  square  by  the  length 
of  the  cask  ;  this  product  multiplied,  or  divided,  by 
the  proper  multiplier,  or  divisor,  in  the  table,  Prob.  I. 
will  give  the  content. 


By  the  Sliding  Ride. 

Find  the  difference  between  the  bung  and  head-dia- 
meter, on  the  inside  of  the  slide  marked  C,  and  oppo- 
site thereto  is,  for  each  variety,*  a  number  to  be  added 
to  the  head -diameter,  for  the  mean  diameter  required. 
Then,  as  the  gauge-point  on  D,  is  to  the  length  on  B  : 
so  is  the  mean  diameter  on  D  to  the  content  on  B. 


*  It  has  been  noticed  before,  that  only  the  first  and  second 
variety  of  casks  are  placed  on  the  rule,  which  is  described  at 
the  beginning-  of  this  chapter.  The  contents  of  casks  are 
generally  found  by  the  sliding-  rule,  for  which  reason  the  rules 
for  finding  the  content  by  the  pen,  are  given  in  such  a  manner 
as  to  agree  therewith.  The  above  methods  of  finding  a  mean 
diameter  are  not  strictly  true.  See  H  dissertation  on  this  sub- 
ject, in  Moss's  Gauging,  sect.  X. 


-'♦IS 


Of  Gauging. 


EXAMPLES. 

1.  Suppose  the  bung-diameter  of  a  cask  to  he  32 
inches,  the  head-diameter  24  inches,  and  the  length 
40  inches  ;  the  content  in  ale  gallons,  for  each  variety. 
is  required? 

1.  For  the  spheroid,  or  first  variety. 


Bung-diameter     32 
Head-diameter    24 

Mean-diameter  29.6 
29.6 

Difference             8 
Multiplier            .7 

5  6 

1776 
2664 
592 

Head-diam.      24. 

876.16  square. 
40  length. 

359.05)35046.40(97.6  gJ 
273190 

218550 

3120 


In  a  similar  manner  you  will  find  the  content  for 
the  second  variety  to  be  94.46  ale  gallons,  for  the 
third  variety  90.87  ale  gallons;  and  for  the  fourth  va- 
riety 88.34  ale  gallons. 

By  the  Sliding  Rule. 

For  the  first  variety  against  8  (the  difference  be- 
tween the  bung  and  head  diameter)  in  the  line  of 
inches,  you  will  find  5.6  on  the  line  marked  spheroid, 


Of  Gauging.  269 

which  added  fo  the  head-diameter  24,  gives  the  mean- 
diameter  29.6.    Then 

onD.      onB.  onD.  on  B. 

18.95     :    40     : :     29.6  mean-diam.     :  97.6 

In  the  same  manner  the  contents  for  the  other  va- 
rieties may  be  found,  by  using  their  respective  mean- 
diameters,  without  removing  the  slider. 

2.  Suppose  the  bung-diameter  of  a  cask  to  be  26.5 
inches,  head-diameter  23  inches,  and  length  28.3 
inches  5  the  content  in  ale  gallons,  for  each  variety,  is 
required  ? 

f  50.8    for  the  first  variety. 
a  s  J  30-°    f°r  *ne  second  variety. 
Jln      J  48.8    for  the  third  variety. 
^48.28  for  the  fourth  variety. 


To  find  the  Contents  of  any  Cask, 

GENERAL   RULE. 

Add  into  one  sum 
39  times  the  square  of  the  bung-diameter, 

25  times  the  square  of  the  head-diameter,  and 

26  times  the  product  of  those  diameters  ; 
multiply  the  sum  by  the  length  of  the  cask,  and  the 
product  by  the  number  .00034;  then  this  last  product 
divided  by  9  will  give  the  wine  gallons,-  and  divided 
by  11  will  give  the  ale  gallons.  * 

Dr.  Hutton's  Diet.  Vol.  I.  page  52S, 


.TO 


Of  Oaugilh 


A  General  Table  for  finding  the  Content  of  any  Cask 
by  the  Sliding  Rule. 


Q. 

, — 

W.  <;. 

A.     G. 

1   Q- 

w.  e. 

A.     G. 

Q. 

w.  ... 

A.      G. 

.50 

.67 

21.95 

.84 

18.36 

20.29 

.51 

23.45 

.08 

19.69 

.85 

18.28 

20.20 

.52 

21.13 

.69 

21.66 

.86 

18.20 

20.12 

.53 

21.04 

.70 

19.51 

21.56 

.87 

18.13 

20.03 

.54 

20.95 

.71 

19.43 

21.47 

.88 

18.05 

19.95 

.55 

20.85 

.72 

19.35 

21.37 

.89 

17.97 

19.86 

.56 

20.76 

92.94 

.73 

19.26 

21.28 

.90 

17.89 

19.77 

.57 

.74 

19.18 

21.18 

.91 

17  81 

19.68 

.58 

20.57 

22.74 

.75 

19.09 

21.09 

.92 

17.74 

19.60 

.59 

20.43 

22.64 

.76 

19.01 

21.01 

|    .93 

17.67 

10.52 

.60 

20.39 

22.54 

.77 

18.93 

20.92 

.94 

17.59 

19.44 

.61 

20.30 

22.45 

.78 

18.85 

20.83 

.95 

17.51 

19.36 

.62 

20.21 

22.35 

.79 

18.77 

20.74 

.96 

17.44 

19.27 

.63 

20.13 

22.25 

.80 

18.69 

20.65 

.97 

17.37 

19.18 

.64 

20.04 

22.15 

.81 

18.61 

20.56 

.98 

17.30 

19.10 

.65 

19.95 

22.05 

.82 

18.53 

20.47 

.99 

17.22 

19.02 

.66 

19.87 

21.86 

.83 

18.44 

20.38 

!  .100 

17.15 

18.95 

The  above  table  of  gauge-points  was  calculated  by 
Mr.  John  Lowry,  an  officer  in  the  excise,  and  an  in- 
genious mathematician.  See  the  Mathematical  and 
Philosophical  Repository,  page  119,  &c. 

The  use  of  the  Table. 

Divide  the  head-diameter  by  the  bung-diameter,  to 
two  places  of  decimals;  find  the  quotient  in  the  co- 
lumn marked  Q,  and  against  it  stand  the  wine  and  ale 
gauge-points?    Then, 

As  the  gauge -point  on  D  :  the  length  of  the  cask 
on  B  :  :  the  bung-diameter  on  D  :  the  content  on  B. 


EXAMPLES. 


1.  The  head-diameter  of  a  cask  is  31.8  inches,  the 
bung-diameter  44.8  inches,  and  the  length  of  the  cask 


Of  Gauging. 


371 


54  inches  $  required  the  content  in  ale  and  wine  gal- 
lons ? 

Bung-diam.  44.8    Head-diam.  34.8  Bung-diam.  44.8 

44.8                         34.8  Head-diam.  34.8 


3584 
1792 
1792 

sum. 
length. 

2784 
1392 
1044 

3584 
1792 
1344 

2007.04 
39 

1211.04  prod. 

25 

1559.04 
26 

1806336 
602112 

605520 
242208 

935424 
311808 

78274.56 
30276 

30276.00 

40535.04 

40535.04 

S050622.4 
.00034 

149085.6 
54 

322024896 
241518672 

5963424 
7454280 

11)2737.211616 

8050622.4 

24S.8374  ale  gallons. 

304.1346  wine  galls. 

By  the  Sliding  Rule. 

The  quotient  arising  by  dividing  the  head-diameter 
by  the  bung-diameter  is  .77,  and  the  wine  and  ale 
gauge-points  are  18.93  and  20.92. 


18.93?        ^ 
20.92  £  °nD: 


tj  jv      C304W.G.onB. 

54  on  B  : :  44.8  on  D  :  |  mA  G    _ 


Of  Gauging. 

2.  The  head-diameter  of  a  cask  is  94.0  inches;  \>umg- 
diameter  31.5,  and  length  iz  inches;  required  its 
content  in  ale  and  wine  gallons  ? 

Jlns    J   95-7'66  a*e  gallons. 
\  117.047  wine  gallons. 

The  content  by  the  sliding  rule  is  exactly  the  same. 

3.  Required  the  content  of  a  cask  in  ale  and  wine 
gallons,  whose  head-diameter  is  24  inches,  bung-diam- 
eter 32  inches,  and  length  40  inches  ? 

q       C 112-28  wine  gallons. 
**ns'  I   91.86  ale  gallons. 

The  content  by  the  sliding  rule  is  the  same. 

4.  The  bung-diameter  of  a  cask  is  48  inches,  head- 
diameter  35.8  inches,  and  the  length  55  inches  5  re- 
quired the  content  in  ale  and  wine  gallons  ? 

-       C  2S3.178  ale  gallons. 
1 346.106  wine  s-allons. 

The  content  is  the  same  by  the  sliding  rule. 

5.  The  head-diameter  of  a  cask  is  28.2  inches,  bung- 
diameter  33.8  inches,  and  length  48  inches  ;  required 
the  content  in  ale  and  wine  gallons  ? 

132.367  ale  gallons. 
61.782  wine  gallons. 


Ms.\lr 


The  content  by  the  sliding  rule  is  the  same. 

PROBLEM    XI. 

Of  the  Ullage  of  Casks. 

The  ullage  of  a  cask  is  what  it  contains  when  only 
partly  filled,  and  is  considered  in  two  positions,  viz, 
standing  on  its  end,  or  lying  on  one  side. 


Of  Gauging.  273 

To  Ullage  a  lying  Cask. 

RULE. 

Divide  the  wet  inches  by  the  bung-diameter :  find 
the  quotient  in  the  column  height,  in  the  table  at  the 
end  of  this  chapter;  take  out  the  corresponding  area 
seg.  multiply  this  area  by  the  content  of  the  cask,  and 
that  product  by  1.27324;  or  divide  it  by  .7854,  the  last 
product,  or  quotient,  will  give  the  ullage  nearly. 

By  the  Sliding  Rule. 

Set  the  bung-diameter  on  C  to  100  on  the  line  mark- 
ed seg.  ly.  or  SL,  viz.  segments  lying :  then  look  for 
the  wet  inches  on  C,  and  observe  what  number  stands 
against  it  on  the  segments,  which  call  a  fourth  num- 
ber. Then  set  100  on  A  to  the  content  of  the  cask 
upon  B,  and  against  the  fourth  number,  before  found, 
on  A  is  the  quantity  of  liquor  in  the  cask  on  B. 

EXAMPLES. 

1.  Supposing  the  bung-diameter  of  a  lying  cask  to 
be  32  inches,  its  content  97.6  ale  gallons  ;  required  the 
ullage  for  19  wet  inches  ? 

1.000  whole  diameter. 
32)19.000(.594  height. 

,7854  whole  area* 

0.406  height.     Area  seg.  .299255 

.486145  rem. 
The  content        97.6 


2916870 
3403015 
4375305 


.7854)47.4477520(64.4  gal. 
Note.     Because,  in  this  example,  the  quotient  of 
the  wet  inches,  divided  by  the  bung-diameter,   exT 


on  C. 

nnSl, 

onC. 

:      100 

:  :      19 

on  A. 

onB. 

on  A. 

As  loo 

:    97.6 

:  :      8&0 

Of  Hanging. 

eeedtthe  heights  in  the  (able,  it  is  necessary    to  sub- 
trait  it  from  the  whole  diameter,  &e. 


Thj  the  Sliding  Hulr. 

on  & 

09.5  fourth  number, 
on  STi. 
:    61  gallons,  .tins. 

2.  The  bung-diameter  of  a  lying  cask  is  20  inches, 
the  content  18.3  ale  gallons  and  the  wet  inches  11 ;  re- 
quired the  quantity  of  liquor  in  the  cask? 

a       $20.46  by  calculation. 
Mns'  1 19.9     by  the  sliding  rule. 

3.  The  bung-diameter  of  a  lying  cask  is  32  inches, 
its  content  91.6  gallons,  and  the  wet  inches  8 ;  required 
the  quantity  of  liquor  in  the  cask? 

a       C  1?.9  gallons,  by  calculation. 
*  US'  £  16.    gallons,  by  the  sliding  rule. 


To  Ullage  a  standing  Cask. 


Multiply  the  difference  between  the  squares  of  the 
bung  and  head  diameters,  by  the  square  of  the  dis- 
tance of  the  liquor's  surface  from  the  middle  of  the 
cask;  and  divide  the  product  by  the  square  of  half  the 
length  of  the  eask,  subtract  one-third  of  the  quotient 
from  the  square  of  the  bung-diameter,  and  multiply 
the  remainder  by  the  distance  of  the  liquor's  surface 
from  the  middle  of  the  cask. 

The  last  product  divided  by  359.05  for  ale,  or  294.12 
for  wine,  will  give  the  quantity  of  liquor  above,  or  un- 
der half  the  content  of  the  cask,  according  as  the  wet 
inches  exceed,  or  fall  short  of  half  the  length  of  the 
eask. 


Of  Ganging. 


By  the  Sliding  Rule. 

Set  the  length  of  the  cask  on  C  to  100  on  the  line 
marked  seg.  st.  or  SS,  viz.  segments  standing ;  then 
look  for  the  wet  inches  on  C,  and  observe  what  num- 
ber stands  against  it  on  the  segments,  which  call  a 
fourth  number. 

Then,  set  100  on  A  to  the  content  of  the  cask  upon 
B,  and  against  the  fourth  number,  before  found,  on  A 
is  the  quantity  of  liquor  in  the  cask  on  B. 

EXAMPLES. 


1.  Suppose  in  the  annexed  cask, 
whose  content  is  97.6  ale  gallons, 
that  the  bung  diameter  EF  be  32 
inches,  head-diameter  AB  24  inch- 
es, length  40  inches,  and  the  wet 
inches  SH  26 ;  what  quantity  of 
ale  is  contained  in  the  caskP  * 


os— 


*\ 


H 


3  2~ 

=EF 

24—AB 

2fc=SH 

32 

24 

20^1H 

64 

96 

6  diff.=SI 

96 

48 

6 

1024 

diff. 

076 

20=IH 

36  square  of  SI 

448 

36 

20 

26S8 

400=IIP 

1344 

quot. 

1024 
13. 

square  of  EF 
44 

0)  16128 

3)40.32 

1010 

56 
6=  SI 

13.44 


359.03)6063.36(16.837  gallons, 


89*  Of  Banging, 

above  half  the  content  of  the.  cask,  viz.  in  ad  EF.  To 
which  add  48.8  gallons,  half  the  content  of  the  cask, 
and  the  gum  is  65.687,  the  quantity  of  the  liquor  in 
the  cask. 

By  the  Sliding  Rule* 


on  C.     on  SS.     on  C. 

on  SS. 

As  40     :     100    : :   26      : 

«6.1  fourth  numher. 

on  A.     on  B     on  A. 

on  B. 

As  100    :    71.6  :  :  66.1     : 

64.5  gallons,           Jlns. 

2.  The  bung-diameter  of  a  standing  cask  is  35  inch- 
es, head-diameter  28.7,  length  40,  wet  inches  30  ;  con- 
tent in  ale  gallons  121.5,  in  wine  gallons  148.5;  re- 
quired the  content  of  the  ullage  iu  ale  and  wine»gal- 
lons?  a       J    93-9-»  ale  gallons. 

*  £  114.76  wine  gallons. 
The  answer  by  the  sliding  rule  is  the  same. 

3.  The  bung-diameter  of  a  standing  cask  is  26.5 
inches,  head-diameter  23  inches,  length  28.3,  wet  inch- 
es 11 ;  and  the  content  in  ale  gallons  48.3 ;  required 
the  ullage? 

a       J  I801  gallons,  by  calculation. 
Jln  '  1 18.3    by  the  sliding  rule. 


The  use  of  the  fallowing  table  in  gauging  has  been 
shewn '.at  the  beginning  of  the  11th  problem ;  but  its 
chief  use  is  for  finding  the  area  of  a  segment  of  a  cir- 
cle.    Thus, 

Divide  the  height  of  the  segment,  by  the  diameter 
of  that  circle  of  .which  it  is  the  segment,  to  three  pla- 
ces of  decimals ;  find  the  quotient  in  the  column  height, 
take,  out  the  corresponding  Area  Seg.  which  multiply 
by  the  square  of  the  aforesaid  diameter,  and  the  pro- 
duct will  be  the  area  of  the  segment  required. 

If  the  quotient  of  the  height  by  the  diameter  be 
greater  than  15,  subtract  it  from  an  unit,  and  find  the 


pf  Gauging,  277 

area  seg.  corresponding  to  the  remainder;  which  sub- 
tract from  .7854  for  the  area  seg. 

If  the  quotient  of  the  height  by  the  diameter  do  not 
terminate  in  three  figures,  find  the  area  seg.  answering 
to  the  first  three  decimals  of  the  quotient;  subtract  it 
from  the  next  greater  area  seg.  multiply  thv*  remainder 
by  the  fractional  part  of  the  quotient,  and  add  the  pro- 
duct to  the  first  area  seg.  taken  out.  —— 


EXAMPLES. 


1.  Required  the  area  of  the  segment  of  a  circle, 
whose  height  is  3.25 ;  the  diameter  of  the  circle  be- 
ing uO  ? 

50)3.25(.065  quotient,  or  tabular  height. 
The  tabular  segment  is  .021659,  which  multiplied  by 
2500,  the  square  of  the  diameter,  gives  54.1475,  the 
area/required. 

™  Required  the  area  of  the   segment  of  a  circle, 
whose  height  is  46.75,  and  diameter  of  the  circle  50? 
\1.000 
50)46.75/  .93.1  quotient.  Whole  area  .7854 

.065  tab.  height.  Area  seg.  .021659 


Remains,  tabular  area  seg.  .763741 
which  multiplied  by  2500  gives  190j.3525,  the  area 
required. 

3.  Required  the  area  of  the  segment  of  a   circle 
whose  height  is  2,  and  diamet  r  of  the  circle  52? 

52)  2.000(.0  8||=. 03  8T6T=  quotient. 
The  area  seg.  answering  to  .038  is  .009763 ;  the  next 
greater  area  seg.  is  .010148;  the  difference  is  .000^85, 
T6T  of  which  is  000177,  which  added  to  .009763  gives 
.009940,  the  area  seg.  corresponding  to  .038T\ :  heuce 
.009940 X square  of  52=26.87776  answer. 


A  a 


Areas  of  the  Segments  of  a  Circle. 


A  TABLE  OF  THE  AREAS 


SEGMENTS  OF  A  CIRCLE, 

Hlwse  Diameter  is  Unity,  and  supposed  to  be  divided 
into  1000  equal  jiarts. 


Beig. 

.bf a  Se$, 

f  J1<'i$- 

.  trea  Meg. 

"V/^-. 

.  iff  a  Seg. 

.001 

.000042 

.036 

.009008 

.071 

.024680 

.002 

.000119 

.037 

.009385 

.072 

.025195 

.003 

.000219 

.038 

.009763 

.073 

.025714 

.004 

.000337 

.039 

.010148 

1  .074 

.026236 

.005 

.000470 

.040 

.01^537 

|  .075 

.026761 

.006 

.000618 

.041 

.010931 

.076 

.027289 

.007 

.000770 

.042 

.011330 

.077 

.027821 

.008 

.000950 

.043 

.011730 

.078 

.028356 

.009 

.001135 

.044 

.012142 

.079 

.028894 
.02940k 

.010 

.001320 

.045 

.012554 

.080 

.011 

.001533 

.046 

.012971 

.081 

.029979 

.012 

.001746 

.047 

.013392 

.082 

.030526 

.013 

.001968 

j  .048 

.013818 

.083 

.0310,  6 

.014 

.002199 

1  .049 

.014247 

.084 

.031629 

.015 

.002438 

j  .050 

.014681 

.085 

.032186 

.016 

.002685 

.051 

.014119 

,086 

.032745 

.017 

.002940 

.052 

.015561 

.087 

.033307 

.018 

.003202 

.053 

.016007 

.088 

.033872 

.019 

.003471 

|  .054 

.016457 

.089 

.034441 

.020 

.003748 

.055 

.016921 

.090 

.035011 

.021 

.004031 

.056 

.017369 

.091 

.035585 

.022 

.004322 

.057 

.017831 

.092 

.036162 

.023 

.004618 

.058 

.018296 

.093 

.036741 

.024 

.004921 

.059 

.018766 

.094 

.037323 

.025 

.005230 

.060 

.019239 

.095 

.037909 

.026 

.005546 

.061 

.019716 

.096 

•038496 

.027 

.005867 

.062 

.020196 

.097 

.039087 

.028 

'  .006194 

.063 

.020691 

.098 

.039680 

.029 

.006527 

.064 

.021168 

.099 

.040276 

.030 

.006865 

.065 

.021650 

.100 

.040875 

.031 

.006909 

.066 

.022154 

.101 

.041476 

.032 

.007558 

.067 

.022652 

.102 

.042080 

.033 

.007913 

.068 

.023154 

.103 

.042687 

.034 

.008275 

.069 

.023659 

.104 

.043296 

.035 

.008638 

.070 

.024168 

.105 

.043908 

Areas  of  the  Segments  of  a  Circle. 


279 


Heig. 

Area  Seg. 

Heig. 

Area  Seg. 

Heig. 

Area  Seg. 

■    — 

— _____ 

__ — 

■ — 

_— . 

.106 

.044522 

.152 

.075306 

.198 

.110226 

.107 

.045139 

.153 

.076026 

.199 

.111024 

.108  . 

.045759 

.154 

.076747 

.200 

.111823 

.109 

.046381 

.155 

.077469 

.201 

.112624 

.110 

.047005 

.156 

.078194 

.202 

.113426 

.111 

.047632 

.157 

.078921 

.203 

.114230 

.112 

.048262 

.158 

.079649 

.204 

.115035 

.113 

.048894 

.159 

.080380 

.205 

.115842 

.114 

.049528 

.160 

.081112 

.206 

.116650 

.115 

.050165 

.161 

.081846 

.207 

.117460 

.116 

.050804 

.162 

.082582 

.208 

.118271 

•117 

.051446 

.163 

.083320 

.209 

.119083 

.118 

.052090 

.164 

.084059 

.210 

.119897 

.119 

.052736 

.165 

.084801 

.211 

.120712 

.120 

.053385 

.166 

.085544 

.212 

.121529 

.121 

.054036 

.167 

.086289 

.213 

.122347 

.122 

.054689 

.168 

.087036 

.214 

.123167 

.123 

.055345 

.169 

.087785 

.215 

.123988 

.124 

.056003 

.170 

.088535 

.216 

.124810 

.125 

.056663 

.171 

.089287 

.217 

.125634 

**26 

.057226 

.172 

.090041 

.218 

.126459 

.127 

.057991 

.173 

.090797 

.219 

.127285 

.128 

.058658 

.174 

.091554 

.220 

.128113 

.129 

.059327 

.175 

.092313 

.221 

.128942 

.130 

.059999 

.176 

.093074 

|  .222 

.129773 

.131 

.060672 

.177 

.093836 

1  .223 

.130605 

.132 

.061348 

.178 

.094601 

I  .224 

.131438 

.133 

.062026 

.179 

.095366 

j  .225 

.132272 

.134 

.062707 

.180 

.096134 

,  .226 

.133108 

.135 

.063389 

.181 

.096903 

.227 

.133945 

.136 

.064074 

.182 

.097674 

.228 

.134784 

.137 

.064760 

.183 

.098447 

.229 

.135624 

.138 

.065449 

.184 

.099221 

.230 

.136465 

.139 

.065140 

.185 

.099997 

.231 

.137307 

.140 

.066833 

.186 

.100774 

.232 

.138150 

.141 

.067528 

.187 

.101553 

.233 

.138995 

.142 

.068225 

.188 

.102334 

.234 

.139841 

.143 

.068924 

.189 

.103116 

.235 

.140688 

.144 

.069625 

.190 

.103900 

.236 

.141537 

.145 

.070328 

.191 

.104685 

.237 

.142387 

.146 

.071033 

.192 

.105472 

.238 

.143238 

.147 

.071741 

.193 

.106261 

.239 

.144091 

.148 

.072450 

.194 

.107051 

.240 

.144944 

.149 

.073161 

.195 

.107842 

.241 

.145799 

.150 

.073874 

.196 

.108636 

.242 

.146655 

.151 

.074589 

.197 

.109430 

.243 

.147512 

eso 


Areas  of  the  Segments  of  a  Circle. 


Hdir. 

.  //•<•(.' 

1  Bag. 

!  590 

Area  8eg. 

Heisr. 

.  Irea  Seg. 

'1  1 

1HJ71 

.189047 

.336 

.231689 

l  19330 

I  ::\n 

.189055  ' 

.337 

634 

.246 

1  1  091 

1  .292 

.190804 

.338 

,  m 

.150953 

.293 

.191775 

.339 

.234546 

.248 

.151816 

.294 

.192684 

.340 

.473 

.249 

.295 

.193596 

.341 

.236421 

.153546 

.296 

.194509  i 

.342 

.237369 

.154412  ' 

.297 

.195422 

.343 

.23  83 18 

.252 

.155280 

.298 

.196337 

.344 

.239268 

.253 

.156149 

.299 

.197252 

.345 

.240218 

.254 

.157019 

.300 

.198168 

.346 

.241169 

.255 

.157890 

.301 

.199085 

.347 

.242121 

.256 

.158762 

.302 

.20000  > 

.348 

.243074 

.257 

.159636 

.303 

.200922 

.349 

.244026 

.258 

.160510 

.304 

.201841 

.350 

.244980 

.259 

.161386 

.305 

.202761 

.351 

.245934 

.260 

.162263 

.306 

.203683 

.352 

.246889 

.261 

.163140 

.307 

.204605 

.353 

.247845 

.262 

.164019 

.308 

.205527 

.354 

.248801 

.263 

.164899 

.309 

.206451 

.355 

.249757 

.264 

.165780 

.310 

.207376 

.356 

.250715 

.265 

.166663 

.311 

.208301 

.357 

.251673 

.266 

.167546 

.312 

.209227  ! 

.358 

.252631 

.267 

.168430 

"  .313 

.210154 

.359 

.253590 

.268 

.169315 

.314 

•211082 

.360 

.254550 

.269 

.170202 

.315 

•212011  | 

.361 

.255510 

.270 

.171089 

.316 

.212940  j 

.362 

.256471 

.271 

.171978 

.317 

.213871 

.363 

.257433 

.272 

.172867 

.318 

.214802 

.364 

.258395 

.273 

.173758 

.319 

.215733  , 

.365 

.259357 

.274 

.174649 

1  .320 

.216666  I 

.366 

.260320 

.275 

.175542 

1  .321 

.217599 

.367 

.261284 

.276 

.176435 

\    .322 

.218533  1 

.368 

•262248 

.277 

.177330 

.323 

.219468 

.369 

.263213 

.278 

.178225 

.324 

.220404  J 

.370 

.264178 

.279 

.179122 

.325 

.221340 

.371 

.265144 

.280 

.180019 

1  .326 

.222277  1 

.372 

.266111 

.281 

.180918 

1  .327 

.223215 

.373 

.267078 

.282 

.181817 

j  .328 

.224154  | 

.374 

.268045 

.283 

.182718 

.329 

.225093 

.375 

.269013 

.284 

.183619 

1  .330 

.226033 

.376 

•269982 

.285 

.184521 

.331 

.226974  1 

.377 

.270951 

.286 

.185425 

.332 

.227915 

.378 

•271920 

.287 

.186329 

.333 

.228858  ! 

.379 

.272890 

.288 

.187234 

!  .334 

.229801  ! 

.380 

.273861 

.289 

.188140 

|  .335 

.230745  | 

.381 

.274832 

~~ 


Areas  of  the  Segments  of  a  Circle.  281 


Heig.   \ 

Area  Seg.  \\   Heig.   \ 

Area  Seg.   | 

Heig.   | 

Area  Seg. 

.382 

.275803  1 

.422 

.315016 

.462 

.344736 

.383 

.276775  I 

.423 

.316004 

.463 

.355732 

.384 

.277748  j 

.424 

.316992 

.464 

.356730 

.385 

.278721  1 

.425 

.317981 

.465 

.357727 

.386 

.279694  1 

.426 

.318970 

.466 

.358725 

.387 

.280668  | 

.427 

.319959 

.467 

.359723 

.388 

.281642  1 

.428 

.320948 

.468 

.360721 

.389 

.282617  | 

.429 

.321938 

.469 

.361719 

.390 

.283592 

.430 

.322928 

.470 

.362717 

.391 

.284568  J 

.431 

.323918 

.471 

.363715 

.392 

.285544 

.432 

.324909 

.472 

.364713 

.393 

.286521  1 

.433 

.325900 

.473 

.365712 

.394 

.287498  | 

.434 

.326892 

.474 

.366710 

.395 

.288476  | 

.435 

.327882 

.475 

.367709 

.396 

.289453 

.436 

.328874 

.476 

.368708 

.397 

.290432 

.437 

.329866 

.477 

.369707 

.398 

.291411 

.438 

.330858 

.478 

.370706 

.399 

.292390 

.439 

.331850 

.479 

.371705 

.400 

.293369 

.440 

.332843 

.480 

.372704 

.401 

.294349 

.441 

.333836 

.481 

.37*3703 

.302 

.295330 

.442 

.334829 

.482 

.374702 

.403 

.296311 

.443 

.335822 

.483 

.375702 

.404 

.297292 

.444 

.336816 

.484 

.376702 

.405 

.298273 

.445 

.337810 

.485 

.377701 

.406 

.299255 

.446 

.338824 

.486 

.378701 

I  .407 

.300238 

.447 

.339798 

.487 

.379700 

.408 

.301220 

.448 

.340793 

.488 

.380700 

.409 

.302203 

.449 

.341787 

.489 

.381699 

.410 

.303187 

.450 

.342782 

.490 

.382699 

.411 

.304171 

.351 

.343777 

.491 

.383699 

.412 

.305155 

.452 

.344772 

.492 

.384699 

.413 

.306140 

.453 

.345768 

.493 
.494 

.385699 

.414 

.307125 

.454 

.346764 

.386699 

.415 

.308110 

.455 

.347759 

.495 

.387699 

.416 

.309095 

.456 

.348755 

.496 

.388699 

.417 

.310081 

.457 

.349752 

.497 

.389699 

.418 

.311068 

j  .458 

.350748 

.498 

.390699 

.419 

.312054 

.459 

.351745 

.499 

.391699 

.420 

.313041 

1  .460 

.352742 

.500 

.392699 

.421 

1  .314029 

.461 

.353739 

a:  a  2 


Ml  Surveying. 

CHAPTER  VI. 

SuRVEriSG. 

SURVEYING  is  the  art  of  measuring,  planning,  and 
finding  the  superficial  content,  of  any  field,  or  parcel 
of  land.  In  this  kind  of  measuring,  th  -  area  or  su- 
perficial content  id  always  expressed  in  acres  ;  or  acres, 
.  and  perches:  and  the  lengths  of  all  lines,  in  the 
field,  or  parcel  of  land,  are  measured  with  a  chain, 
such  as  is  described  at  page  78. 

A  line,  or  distance  on  the  ground,  is  thus  measured. 
Having  procured  to  small  arrows  or  iron  rods,  to  stick 
in  the  ground  at  the  end  of  each  chain  ;  also  some  sta- 
tion-staves, or  long  poles  with  coloured  flags,  to  set  up 
at  the  end  of  a  station-line,  or  in  the  angles  of  a  field  \ 
two  persons  take  hold  of  the  chain,  one  at  each  end ; 
the  foremost,  for  the  sake  of  distinction,  is  called  the 
leader,  the  hindermost  the  follower. 

A  station-staff  is  set  up  in  the  direction  of  the  line  to 
he  measured,  if  there  be  not  some  ohject,  as  a  tree,  a 
house,  &c.  in  that  direction. 

The  leader  takes  the  10  arrows  in  the  left  hand,  and 
one  end  of  the  chain,  by  the  ring,  in  his  right  hand, 
and  proceeds  towards  the  station -staff,  or  other  object. 
The  follower  stands  at  the  beginning  of  the  line,  hold- 
ing the  other  end  of  the  chain,  by  the  ring,  till  it  is 
stretched  straight,  and  laid,  or  held  level,  by  the  lead- 
er, whom  he  directs,  by  waving  his  hand  to  the  right  or 
left,  till  he  see  him  exactly  in  a  line  with  the  object 
towards  which  they  are  measuring.  The  leader  then 
sticks  an  arrow  upright  in  the  ground,  as  a  mark  for 
the  follower  to  come  to,  and  proceeds  forward  another 


Surveying;. 


288 


chain,  at  the  end  of  which  he  is  directed,  as  before,  by 
the  follower  ;  or  he  may  now,  and  at  the  end  of  every 
other  chain,  direct  himself,  by  moving  to  the  right  or 
left,  till  the  follower  and  the  object  measured  from,  be 
in  one  straight  line.  Having  stuck  down  an  arrow, 
as  before,  the  follower  takes  up  the  arrow  which  the 
leader  first  stuck  down.  And  thus  they  proceed  till 
all  the  10  arrows  are  employed,  or  in  the  hands  of  the 
follower,  and  the  leader,  without  an  arrow,  is  arrived 
at  the  end  of  the  eleventh  chain  length.  The  follow- 
er then  sends  or  carries  the  10  arrows  to  the  leader, 
who  puts  one  of  them  down  at  his  end  of  the  chain, 
and  proceeds  with  the  other  nine  and  the  chain  as  be- 
fore. The  arrows  are  thus  changed  from  the  one  to 
the  other,  till  the  whole  line  is  finished,  if  it  exceed 
10  chains ;  and  the  number  of  changes  shews  how  ma- 
ny times  10  chains  the  line  contains.  Thus,  if  the 
whole  line  measures  36  chains  45  links,  or  3645  links, 
the  arrows  have  been  changed  three  times,  the  follower 
will  have  5  arrows  in  his  hand,  the  leader  4,  and  it  will 
he  45  links  from  the  last  arrow,  to  be  taken  up  by  the 
follower,  to  the  end  of  the  line. 

Of  the  Surveying-Cross,  or  Cross-Staff. 


The  surveying  cross  consists  of 
two  pair  of  sights  at  right  an- 
gles to  each  other :  these  sights 
are  sometimes  pierced  out  in  the 
circumference  of  a  thick  tube  of 
brass;  and  sometimes  the  cross- 
staff  consists  of  four  sights 
strongly  fixed  upon  a  brass  cross, 
and  when  used  is  screwed  on  a 
staff,  having  a  sharp  point  to 
stick  in  the  ground.  The  accu- 
racy of  the  cross-staff  depends  on 
the  sights  being  exactly  at  right 


2S*  Surveying. 

angles  to  each  oilier.  A  cross-slaff  may  be  easily 
made  by  any  carpenter.  Thus,  take  a  piece  of  beech 
or  box,  ADBC,  of  tour  or  live  inches  in  breadth,  and 
three  or  four  indies  in  depth,  and  upon  ADBC  draw 
two  lines,  AB  ai\d  CD,  crossing  each  other  at  right 
angles.  Then  wilh  a  fine  saw  make  two  slits,  ABG 
and  CDH,  of  about  two  inches  in  depth  ;  fix  this  piece 
of  wood  upon  a  staff  8,  of  about  4  and  a  half  or  5  feet 
in  length,  pointed  at  one  end,  so  that  it  may  easily 
stick  into  the  ground. 


PROBLEM    I. 

To  Measure  Off-Sets  with  a  Chain  and  Cross-Staff. 

Let  Abcdefg  be  a  crooked  hedge,  river,  or  brook, 
&c.  and  AG  a  base  line.  First,  begin  at  the  point  A, 
and  measure  towards  G  :  when  you  come  to  B,  where 
you  judge  a  perpendicular  must  be  erected,  place  the 
cross-staff  in  the  line  AG  in  such  a  position  that  both 
G  and  A  may  be  seen  through  two  of  the  sights,  look- 
ing forward  towards  G,  and  backward  towards  A. — 
Then  look  along  one  of  the  cross-sights,  and  if  it  point 
directly  to  the  corner,  or  bend  at  b,  the  cross-staff  is 
placed  right;  otherwise  move  backward  or  forward  a- 
long  AG  till  the  cross-sights  do  point  to  6,  and  measure 
B6,  which  set  down  in  links:  proceed  thus  till  you  have 
taken  all  the  off-sets,  as  in  the  following 


Surveying. 


285 


FIELD-BOOK. 


Off-sets 
left. 

Base  line  AG, 
or,  O,  stations. 

Off-sets 
right. 

62 

84 
70 
98 
57 
91 

OA. 

43 
220 
340 
510 
634 
785 



dS 


To  lay  down  the  Plan. 

Draw  the  line  AG  of  an  indefinite  length.  Then, 
by  a  diagonal  scale,  such  as  described  at  page  48,  set 
oft'AB  equal  to  45  links,  draw  B6  perpendicular  to  AG, 
and  equal  to  62  links.  Next  set  off  AC  equal  to  220 
links,  or  2  chains,  20  links  ;  draw  Cc  perpendicular  to 
AG,  and  equal  to  34  links :  then  set  off  AD  equal  to 
340  links;  or  3  chains  40  links,  and  make  Dd  equal  to 
70  links  :  proceed  thus  till  you  have  completed  the 
figure. 

To  cast  up  the  Content. 

AB6  must  be  measured  as  a  triangle  ;  BCcfr,  CDc?c, 
DEec?,  &c.  must  be  measured  as  trapezoids;  see  page 
93.  Some  authors  direct  you  to  add  all  the  perpen- 
diculars B6,  Cc,  &c.  together,  and  divide  their  sum  by 
the  number  of  them,  then  multiply  the  quotient  by  the 
length  AG;  but  this  method  is  always  erroneous,  ex- 
cept the  off-sets  B6,  Cc,  &c.  be  equally  distant  from 
each  other. 


286 


Surveying 


bs> 

3 

1 

■  ->  1 1    1 

II  II 

a* 

«£           ©    CO    1 

05    *> 

1 

O     1 

O    1 

to  cm 

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oca 

do 

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to 
cm 

n   = 

II  3 

H       II 

n 

11  1 

6SO 
II   II 

CM 

►*•   H* 

►*• 

to 

Cx 

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00   3s 

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blc  area  of  A 

3ft. 

25550 

sa  doi 

bl e  area  of  B 

Ccb. 

48480 

=  doi 

ihle  area  of  C 

ndc. 

28560 

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ihlearea  of  D 

Red. 

19220 

=  doi 

bio  area  of  E 

Ffi. 

22348 

=  doi 

lble  area  of  F< 

%/■ 

2)116948  =  double  area  of  the  whole  in  sq.  links. 
58474  =  area  in  square  links. 
.58474  =  area  in  acres =0  A.  2  R.  13.5584  P. 


2.  Required  the  plan  and  content  of  part  of  a  field^ 
from  the  following  field-book. 


Surveying. 


2S7 


Off-sets 
left. 

Base  line 
AG. 

Off-sets  right. 

60 
150 

0A. 

100 
250 
325 
450 
550 
700 

100 

60 

150 

46 

Cross  Hedge. 

The  figure  must  be  laid  down,  and  the  content  cal- 
culated as  in  the  first  example.  Thus  you  will  find 
the  area  of  the  part  E  d  c  b  a  A  to  be  1  R.  13.  4.  P. 
and  of  the  part  E/g  G  to  be  30  perches ;  so  that  the 
whole  is  2  roods  3.4  perches. 


PROBLEM    II. 


To  measure  a  Field  in  the  form  of  a  Trapezium. 

Set  up  station-staves,  or  long  poles,  at  the  corners 
A,  B,  C.  Then  begin  at  D,  and  measure  along  the 
diagonal  DB,  in  a  right  line,  till  you  come  to  the  place 
of  the  perpendicular  AF,  which  will  be  known  by 
looking  backward  and  forward  through  the  sights  of 
the  cross-staff,*  as  before  directed.  Make  a  mark  at 
F,  and  measure  the  perpendicular  AF ;  then  proceed 
from  F  towards  B,  and  when  you  come  to  E,  the  place 
where  the  second  perpendicular  will  fall,  make  a  mark, 
and  measure  the  perpendicular  EC :  lastly,  continue 
your  measure  from  E  to  B.  You  may  either  draw  a 
rough  plan  of  the  field  by  the  eye,  and  write  the  length 
of  the  diagonal  and  perpendiculars  on  it,  which  some 


233 


Survcj/iui 


writers  recommend  as  the  best  method;  or  set  them 
down  in  a  field-hook,  thus : 


Offsets 
'left. 

S Id  lion, 
or  base  line. 

Off-sets 
right. 

849 

on 

600 
1100 
1360 

620 

To  lay  down  the  Plan. 

Draw  the  station-line  DB  equal  to  1360  links,  or 
13.60  chains;  from  1)  set  oft'  I)F  equal  to  600  links, 
or  8  chains  ;  draw  AF  perpendicular  to  DB,  and  equal 
to  o42  links,  or  3  chains  42  links:  make  DE  equal  to 
1190  links,  and  at  E  erect  the  perpendicular  EC  equal 
to  620  links.  Join  DA,  Aii,  BC  and  DC,  and  the 
field  is  constructed. 

CALCULATION. 

This  field  being  a  trapezium,  its  content  must  be 
found  as  directed  in  section  VI.  chapter  1.  part  II. 


Surveying 


289 


AF=r342  links. 
EC=625 

Sum  967 

680  half  the  diagonal  DB. 


77380 
5S02 


6.57560  area  in  acres:=:6  acres  2  rods 
12.096  perches. 


2.  Required  the  plan  and  content  of  a  field  from  the 
following;  field-book. 


Off-sets 
left. 

Stations, 
or  base  lines. 

Off-sets 
right. 

306 

0D. 

214 

362 
592 

210 

Jlnswer.  The  content  is  1  acre  2  roods,  4.3776  per. 

Note.  In  either  of  the  above  figures,  if  the  sides 
DC,  CB,  BA,  AD,  and  the  diagonal  DB  had  b  en 
measured  with  the  chain,  the  figures  might  have  been 
planned,  and  their  contents  found  by  the  rule,  page 
84,  without  using  a  cross-staff,  or  measuring  the^M^ 
pendiculars  AF  and  CE. 

Some,  who  pretend  to  measure  land,  always  measure 
round  a  four-sided  field,  and  cast  up  the  content  by 
adding  every  two  opposite  sides  together,  and  taking 
half  their  sum ;  and  then  multiply  these  half  sums 
into  each  other.  It  may  be  necessary  to  inform  the 
learner,  that  this  method  is  very  erroneous,  and  ought 
never  to  be  practised. 

Bb 


Z90 


Surveying. 


PROBLEM    III. 


To  measure  a  four-sided  Field  with  crooked  Hedges. 

Set  ii|)  staves,  or  poles,  at  the  corners  D,  C,  B,  as 
before  directed.  Then  begin  at  A,  and  measure,  from 
AB,  noting  all  the  necessary  oft-sets,  and  in  this  man- 
ner go  round  the  field,  then  measure  the  diagonal  AC. 


FIELD-BOOK. 


V 

Stations, 
or  base  lines. 

Off-sets 
right. 

OA. 

300 
400 
550 
750=AB 

100 

130 

80 



OB. 

200 
700 

1500=BC 

200 
150 

oc.     • 

300 
500 
700 
900 

iooo=CD 

200 

100 

150 

50 

150 
200 
100 

OD. 

200 
400 
700 
soozrDA 

Diagonal  AC =1294. 

Surveying.  29.1 


CALCULATION. 

Acres. 
The  sides  of  the  triangle  ABC  are  750,  "J 

1500,  and  1291  links,  and  its  content,  by  I  4.85239 

the  rule,  page  84  J 

The  sides   of  the  triangle  ACD  are  1000,1 

800,  and  1294  links,  and   its  content,  by  I  3.99907 

the  rule,  page  84  - ' J 

Content   of  the  oft-sets   along  AB= 0.50250 

Ditto : along  BC= 1.67500 

Ditto along  CD=  1.07500 


Sum=  12.10396 
Content  of  the  off-sets  along  DA,  deduct       1.00000 

Remains  the  content  of  the  field=li.l0396 


PROBLEM    IV. 

How  to  Measure  an  Irregular  Field. 


a' 


The  way  to  measure  irregular  land,  is  to  divide  it 
into  trapeziums  and  triangles,  thus : 

First,  look  over  the  field,  and  set  up  marks  at  every 
angle,  and  by  those  marks  you  may  see  where  to  have 
a  trapezium,  as -ABCI  in  the  following  figure. 


Muvveyiug> 


H<< 


_ b* "^ 


\r 


WAG        %         .S/ 


/v- 


I 


!5» 


hi  begin  and  measure  in  a  direct  line  from  A  to- 
warils  C  5  but  when  you  come  to  a>  set  up  your  cross, 
and  try  whether  you  be  in  a  square  to  I  (as  is  before 
directed)  :  and  then  measure  the  perpendicular  a  I, 
which  is  482  links;  then  measure  forward  again  to- 
ward C,  but  when  you  come  to  b,  set  up  yoar  cross, 
and  try  whether  you  be  in  the  place  where  the  perpen- 
dicular will  fall  5  then  measure  the  perpendicular  6B, 


4 


Surveying.  293 

which  is  206  links ;  then  continue  your  measure  to  C, 
and  you  will  find  the  whole  diagonal  942  links. 

Then  proceed  to  measure  the  trapezium  CDHI,  he- 
ginning  at  C,  and  measuring  along  the  diagonal  line 
towards  H  :  but  when  you  come  to  rf,  set  up  your  cross, 
and  try  if  you  be  in  the  place  where  the  perpendicu- 
lar will  fall :  measure  the  perpendicular  rfD,  which  is 
146  links ;  and  then  measure  forward  till  you  come  to 
c,  and  there,  with  your  cross,  try  if  you  be  right  in  the 
place  where  the  perpendicular  will  fall,  and  measure 
the  perpendicular  el,  which  is  3  chains;  and  from  c 
continue  your  measure  to  H,  and  you  will  find  the 
whole  diagonal  1236  links. 

• 

Then  proceed  to  measure  the  trapezium  HGED, 
beginning  at  H,  and  measuring  along  the  diagonal  line 
towards  E;  but  when  you  come  to  /,  try  with  your 
cross  if  you  be  in  the  place  where  the  perpendicular 
will  fall ;  and  measure  the  perpendicular  /G,  which 
is  448  links;  then  continue  on  your  measure  from/ 
till  you  come  to  «•,  and  there  try  if  you  be  in  a  square 
with  the  perpendicular  gD  ;  and  measure  the  said 
perpendicular,  which  is  294  links ;  then  measure  on 
from  g  to  E,  and  you  will  find  the  whole  diagonal  to 
be  1144  links* 

Then  measure  the  triangle  EFG,  beginning  at  E, 
and  measuring  along  the  base  EG,  till  you  come  to  h, 
and  there  with  your  cross  try  if  you  be  in  the  place 
where  the  perpendicular  will  fall ;  and  measure  the 
perpendicular /iF,  which  is  314  links,  continue  your 
measure  to  G,  and  you  will  find  the  whole  base  to  be 
912  links;  so  you  have  finished  your  whole  field. 

But  to  draw  a  planiof  the  field,  it  would  be  neces- 
sary to  measure  a  few  more  lines,  or  mark  the  points 
$,  6,  &c. 

b  b2 


l\)h 


Of  Sunrffh}^ 


l  ION. 


The  area  of  the  trapezium  ABCI =324043 

The  area  of  the  trapezium  CDH1 

Thfl  area  of  the  trapezium  HGED=420714 

The  area  of  the  triangle  EFG        =14.;  1 8 1 


The  area  of  the  whole      -     1163574 

Cut  off  five  figures  from  the  right-hand,  and  the  re- 
sult will  he  11.63574  acres=  11  acres  2  roods  21.71S4  p. 

2.  Required  the  plan  and  content  of  an  irregular 
field,  from  the  following 


FIELD-ROOK. 


Off-sets 

Station,  or 

Off-sets 

left. 

Base  Line. 

right. 

OK. 

380 

240 

, 



380 

480 

553 

/  35 

265 

. 

940 

600 

140 

1100 





1410 

435 

— — — 

1620 

This  field,  when  constructed,  is  exactly  similar  to 
the  former  one.  The  content  of  the  off-sets  must  be 
found  as  in  problem  1.  Thus, 

Square  Links. 
The  content  of  ABCDEF  will  he  found  601000 
The  content  of  AIHGF 439390 


The  area  of  the  whole     -      1040390 


Cut  off  five  figures,  and  the  result  will  be  10.40390 
acres  =  10  A.  1  11.  24.624  P. 


Practical  Questions.  295 

problem  v. 

To  cut  off  from  a  Plan  a  given  number  of  Acres,  Sft\ 
by  a  line  drawn  from  any  point  in  the  side  of  it. 

Let  A  be  the  given  point  in 
the  annexed  plan,  from  which 
a  line  is  to  be  drawn  towards 
B,  so  as  to  cut  off  5  acres  2 
roods  14  perches.  Draw  AB, 
so  as  to  cut  off  a  quantity  ABC, 
as  near  the  quantity  proposed 
as  you  can  judge  ;  and  suppose 
the  true  quantity  of  ABC, 
when  calculated  to  be  only  4  A.  3  R.  20  P.  which  is 
3  R.  34  P. =114  perches =71250  square  links  too  little. 
Then  measure  AB,  which  suppose  equal  to  1234  links, 
by  the  half  of  which,  viz.  617  links,  let  71250  links 
be  divided ;  the  quotient,  115  links,  will  be  the  alti- 
tude of  the  triangle  to  be  added,  whose  base  is  AB, 
therefore  make  BD=115  links,  and  draw  AD,  which 
will  cut  off  the  quantity  required. 


CHAPTER  VIL 

Practical  (Questions  in  Measuring. 

Question  1.  IF  a  pavement  be  47  feet  9  inches  long, 
and  18  feet  6  inches  broad,  I  demand  how  many  yards 
are  contained  in  it?  Jins.  98  yards  1  foot. 

Quest.  2.  There  is  a  room,  whose  length  is  21.5  feet, 
and  the  breadth  17.5  feet,  which  is  to  be  paved  with 


196  Practical  Questions. 

stoii< s,  each  10  inches  square ;  1  demand  how  many 
such  stones  will  pave  it? 

Jlns.  167|  stones. 

Quest.  3.  There  is  a  room  109  feet  9  inches  about, 
and  9  feet  3  inches  high,  which  is  all  (except  two  win- 
dons,  each  6  feet  6  inches  high,  and  5  feet  i)  inches 
broad)  to  be  hung  with  tapestry  that  is  ell-hroad ;  I 
desire  to  know  how  many  yards  will  hang  the  said 
room  ? 

From  the  content  of  the  room,  subtract  the  content 
of  the  windows,  and  divide  the  remainder by  the  square 
feet  in  a  yard  of  tapestry,  viz.  i4\  feet,  and  you  will 
find  the  answer  to  be  83.59  yards. 


Jlns. 


Quest.  4.  If  the  axis  of  a  globe  be  27.5  inches ;  I 
demand  the  content  solid  and  superficial  r 

10889.24375  inches=6.3  feet  solid. 
2373.835  inches—  1 6.49  feet  superficies. 

Quest.  5.  There  is  a  segment  of  a  globe,  the  diam- 
eter of  whose  base  is  24  inches,  and.  its  altitude  19 
inches  ;  what  is  the  content  solid  and  superficial 
including  the  area  of  its  base  ? 

a       J  .'785.552  inches  the  solidity. 
'  ^  1218.9408  inches  the  superficies. 

The  diameter  of  the  whole  sphere  may  be  found  as- 
in  page  118,  to  be  24.4  inches. 

Quest  6.  If  a  tree  girt  18  feet  6  inches,  and  be  24 
feet  long,  how  many  tons  of  timber  are  contained  in 
that  tree,  using  the  customary  method  of  measuring, 
and  allowing  40  feet  of  timber  to  a  ton  ? 

Jlns.  12  tons  33  feet  4  inches  6  parts. 


#. 


Practical  Questions.  297 

Quest.  7.  There  is  a  cellar  to  be  dug  by  the  floor,  the 

length  of  which  is  S3  feet  7  inches,  and  the  breadth 

18  feet  9  inches,  and  its  depth  is  to  be  5  feet  9  inches ; 

I  demand  how  many  floors  of  earth  are  in  that  cellar  ? 

Ms.  11  floors  56  feet  8  inches  5  parts. 

Note.  That  18  feet  square  and  a  foot  deep  is  a  floor 
of  earth,  that  is,  324  solid  feet. 

Quest.  S.  There  is  a  roof  covered  with  tiles,  whose 
depth  on  both  sides  (with  the  usual  allowance  at  the 
eaves)  is  35  feet  6  inches,  and  the  length  48  feet  9 
inches ;  how  many  squares  of  tiling  are  contained  in 
it?  Jins.  17  squares  30  feet  7\  inches. 

Quest  9.  There  is  a  cone,  the  diameter  at  the  base 
being  42  inches,  and  the  perpendicular  height  94 
inches ;  and  it  is  required  to  cut  off  two  solid  feet 
from  the  top  end  of  it  5  I  demand  what  length  upon 
the  perpendicular  must  be  cut  oft'? 

First,  find  the  solidity  of  the  whole  cone,  43410.628S 
cubic  inches,  or  25.1219  feet,  and  the  cube  of  the  alti- 
tude 830584  inches.  Then,  since  all  similar  solids 
are  in  proportion  to  each  other  as  the  cubes  of  their 
like  parts, 

Feet.  Inches.        Feet.        Inches. 

As  25.1219  :  830584  : :  2  :  66124.297,  the  cube 
of  the  altitude  to  be  cut  oft",  the  cube-root  of  which  is 
40.436  isches,  Answer. 

* 
Quest.  10.  If  a  square  piece  of  timber  be  12  feet 
long,  and  if  the  side  of  the  square  of  the  greater  base 
be  21  inches,  and  the  side  of  the  square  of  the  lesser 
base  be  3  inches  5  how  far  must  I  measure  from  the 
greater  end,  to  cut  off"  five  solid  feet  ? 

You  will  find  the  length  of  the  whole  pyramid  by 
the  rule,  page  118,  to  be  14  feet.    Then  find  the  solid 


Practical  Questions. 

content  of  the  whole  pyramid  14*294     feet,  from  which 
deduct  r>  feet,  the  remainder  is  9.291;  feet. 

Solidity.      Cube  length.     Solidity.    Cube  length. 
As  14.29  if.       :     .2744       ::       9.29 1^        :y    1784 
Or  42.875         :       2744       : :     27.873         :       1784    the 
cube-root  of  which   is   12.128  feet,  which  subtracted 
from  the  whole  length,  14  feet,  leaves  1.872  i'eet^  the 
length  of  5  solid  feet  at  the  greater  enjj. 

Quest.  11.  Three  men  bought  a  grindstone  of  40 
inches  diameter,  which  cost  20  shillings ;  of  which 
sum  the  first  man  paid  9  shillings,  the  second  6  shil- 
lings, and  the  third  5  shillings :  I  demand  how  much 
of  the  stone  each  man  must  grind  down,  proportion- 
able to  the  money  he  paid  ? 

AH  circles  are  to  each  other  as  the  squares  of  their 
diameters  ;  and  each  man  must  grind  away  a  surface 
proportionable  to  the  money  he  paid. 

20  :  20X20  ::  5  :  100,  the  square-root  of  10, 
20  :  20X20  :  :  6  :  120 

Sum  220,  the  square-root  is 
14.832397,  the  radius  which  two  men  must  grind  down 
from  the  centre,  from  which  take  10,  the  radius  which 
one  man  must  grind  down,  there  remains  4.S32397,the 
breadth  of  the  ring  which  the  second  man  must  grind 
away :  from  the  whole  radius  20  subtract  14.832397, 
the  radius  which  two  men  must  grind  away;  the  re- 
mainder is  5.167603,  the  breadth  of  the  ring  the  first 
man  must  grind  away. 


Practical  Questions. 


299 


OR  GEOMETRICALLY  THUS  : 


First,  upon  the  centre  O  describe  the  circle  ABCD, 
and  cross  it  at  right  angles  with  the  two  diameters 
AB  and  CD :  then  divide  the  semidiameter  A  ©  in 
proportion  to  9s.  6s.  and  5s.  the  several  sums  paid  by 
the  three  men  ;  viz.  make  AE  9,  EF  6,  and  F  O  ft:: 
then  divide  EB  into  two  equal  parts  in  d,  and  upon  d9 
as  a  centre,  describe  the  semicircle  EoB;  divide  FB 
into  two  equal  parts  in  c,  and  upon  c,  as  a  centre,  with 
the  radius  e  F,  describe  the  semicircle  F  b  B ;  which 
will  divide  the  semidiameter  ©  C  into  three  such  parts 
as  the  stone  ought  to  be  divided;  and  circles  described 
through  these  points,  will  shew  how  much  each  man 
must  grind  for  his  share. 

This  construction  is  derived  from  the  above  me- 
thod of  calculation  for  B©=20,  QF—5,  and  ©E=ll, 
by  the  property  of  the  circle  B©X©F=©&a,  and 
B*©X©E=©«2;  also  ©«— ©&=«&  and  ©C— Qa=aQ : 
hence  ©&=lo,  a&=4.83,  and  aC=5.l7  nearly. 


.loo  "Practical  ({usstiotu. 

({ifst.  i  j,  A  gard'ner  had  an  upright  cone, 
Out  of  which  should  he  cut  him  a  rolling-stone, 

The  biggest  thai  e'er  it  could  make: 
The  mason  he  said,  that  there  was  a  rule 
For  such  sort  of  work,  but  he  had  a  thick  skull: 
Now  help  him  for  pity'*  sal 
Jlns.  It  must  he  cut  at  one-third  part  of  the  altitude. 

Note.  This  is  property  a  question  in  fluxions,  and 
not  dependant  upon  any  rule  in  this  book.  S  e  Simp- 
son's Geometry,  on  the  Maccima  and  Minima  of  geo- 
metrical quantities,  theorem  XXX. 

Qu<>r,t.  13.  There  is  a  cistern,  whose  depth  is  seven 
tenths  of  the  width,  and  the  length  is  6  times  the 
depth,  and  the  solid  capacity  is  367.3  feet;  I  demand 
the  depth,  width,  and  length,  and  how  many  bushels 
of  corn  it  will  hold  ? 

First,  you  must  find  three  numbers  in  proportion  to 
the  depth,  width,  and  length  thus  :  suppose  the  width 
1,  the  depth  will  be  .7,  and  the  length  4.2;  hence  the 
solidity  will  be  2.94  feet.  But  solids  arc  to  each  other 
as  the  cubes  of  their  like  parts,  consequently 

2.94  feet  :  1  cube  width  ::  367.5  feet  :  125  the  cube 
of  the  real  width,  the  cube-root  of  which  is  0  ('  et 
the  width;  hence  the  depth  is  3.5  feet,  and  length  21 
feet. 

The  content  is  295  bushels  1  peck  4  pints. 

Quest.  14.  Suppose,  sir,  a  bushel  be  exactly  round, 
Whose  depth  being  measured,  8  inches  is  found ; 
If  the  breadth  IS  inches  and  half  you  discover, 
This  bushel  is  legal  all  England  over. 
But  a  workman  would  make  one  of  another  frame  ; 
Sev'n  inch  and  a  half  must  be  the  depth  of  the  same; 
Now,  sir,  of  what  length  must  the  diameter  be, 
That  it  may  with  the  former  in  measure  agree  ? 

Ans.  19.107  inches  must  be  the  diameter  when  the 
depth  is  7'£  inches. 


Practical  Questions.  301 

Quest.  15.  In  the  midst  of  a  meadow,  well   stored 

with  grass, 
I  took  just  an  acre,  to  tether  my  ass : 
How  long  must  the  cord  be,  that,  feeding  all  round, 
He   may'nt   graze   less  nor   more    than   his   acre  of 

ground?  Jlns.  117  feet  9  inches, 

Quest.  16.  A  malster  has  a  kiln,  that  is  16  feet  6 
inches  square;  but  he  is  minded  to  pull  it  down,  and 
build  a  new  one,  that  may  be  big  enough  to  dry  three 
times  as  much  at  a  time  as  the  old  one  will  do ;  I  de- 
mand how  much  square  the  new  one  must  be  ? 

Jlns.  The  side  of  the  new  one  must  be  28  feet  and 
near  7  inches,  or  28.578  feet. 

Quest.  17.  If  a  rfcund  cistern  be  26.3  inches  diame- 
ter, and  52.5  inches  deep;  how  many  inches  diameter 
must  a  cistern  be  to  hold  twice  the  quantity,  the  depth 
being  the  same  ?  and  how  many  ale  gallons  will  each 
cistern  hold  ? 

Jlns.  The  diameter  of  the  greater  cistern  is  37.19 
inches,  and  its  content  202.275  gallons ;  hence  the  con- 
tent of  the  less  cistern  must  be  101.137  gallons. 

Quest.  18.  If  the  diameter  of  a  cask  at  the  bung  be 
32  inches,  and  at  the  head  25  inches,  and  the  length 
40  inches;  how  many  ale  gallons  are  contained  therein  ? 

Jlns.  94.41  gallons,  by  the  general  rule,  page  269. 

Quest.  19.  There  is  a  stone,  20  inches  long,  15  inches 
broad,  and  8  inches  thick,  which  weighs  217  pounds ; 
I  demand  the  length,  breadth,  and  thickness  of  another 
of  the  same  kind  and  shape,  which  weighs  1000 
pounds  ? 

The  cube  of  20,  the  length,  is  8000.  Then 

217  :  8000  : :  1000  :  36866.3594,  whose  cube  root 
is  33.282  inches,  the  length  of  the  stone  weighing 
1000  pounds.    Then  say, 

Ce 


inches. 


302  Practical  questions. 

20  :  .;.;.JsJ  ::    10  :  21.901 
20  :  J.J.2S2  :  :     8  :    18, 
("The  length  sjl 

Ms.  <  The  breadth  jj.'.kh  li 

I^The  thickness        13.31 2  J 

Quest.  20.  If  an  iron  bullet,  whose  diameter  is  4 
inches,  weighs  9  pounds;  what  will  he  the  weight  of 
another  bullet  (of  the  same  metal)  whose  diameter  i- 
9  inches  ?  Jins.  102.515  pounds. 

Quest,  21.  There  is  a  square  pyramid  of  marble, 
each  side  of  its  base*  is  5  inches,  and  the  height  15 
inches,  and  its  weight  is  12  pounds  and  a  quarter;  I 
demand  the  weight  of  another  like  square  pyramid, 
each  side  of  whose  Jjase  is  30  inches? 

The  cube  *of  U  is  125,  and  the  cube  of  30  is  27000. 
Then  (by  End.  XII.  12.)  • 

lb.  lb. 

125  :  12.25  :  :  27000  :  2646 

Ms.  The  weight  is  2646  pounds. 

Quest.  22.  There  is  a  ball  or  globe  of  marble,  whose 
diameter  is  6  inches,  and  its  weight  11  pounds;  what 
will  be  the  diameter  of  another  globe  of  the  same 
marble,  that  weighs  500  pounds  ?         Ms.  21.4  inches. 

Quest.  23.  There  is  a  frustum  of  a  pyramid,  whose 
bases  are  regular  octagons ;  each  side  of  the  greater 
base  is  21  inches,  and  each  side  of  the  less  base  is  9 
inches,  and  its  perpendicular  length  is  15  feet;  I  de- 
mand how  many  solid  feet  are  contained  in  it  ? 

Ms.  119.2  feet* 

Quest.  24.  There  is  a  frustum  of  a  cone,  the  diame- 
ter of  the  greater  base  is  36  inches,  and  the  diameter 
of  the  less  base  is  20  inches,  and  the  length  or  height 
is  215  inches  ;  I  demand  the  length  and  solid  content 
of  the  whole  cone,  and  also  the  solid  content  of  the 
^iven  frustum  ? 


Practical  Questions*  303 

First,  find  the  length  of  the  whole  cone,  thus : 

As  8,  the  difference  between  the  radii  of  the  two 
ends  :  215,  the  length  of  the  frustum  :  :  18,  the  radius 
of  the  greater  end  :  483.75  inches,  the  whole  length  of 
the  cone. 

The  solidity  of  the  whole  cone  is  94.98  feet. 

The  solidity  of  the  frustum  is         78.7  feet. 

quest.  25.  If  the  top  part  of  a  cone  contains  26171 
solid  inches,  and  200  inches  in  length,  and  the  lower 
frustum  thereof  contains  159610  solid  inches  ;  I  de- 
mand the  length  of  the  whole  cone,  and  the  diameter 

of  each  base  ? 

Inches. 

("The  length  of  the  whole  cone  384.3766 

Arts.  \  The  diameter  of  the  greater  base  42.9671 

I  The  diameter  of  the  less  base  22.3568 

Quest.  26.  There  is  a  frustum  of  a  cone,  whose  solid 
content  is  20  feet,  and  its  length  12  feet  ;  and  the 
greater  diameter  bears  such  proportion  to  the  less  as  5 
to  2 ;  I  demand  the  diameters  ? 

First,  I  find  the  content  of  a  conical  frustum,  whose 
diameters  are  5  and  2,  and  depth  12  feet,  to  be  122.5224 
feet.  Then,  as  122.522  :  20  feet : :  25,  square  of  the 
greater  diameter  :  4.080886,  the  square  root  of  which 
is  2.02012  feet,  the  true  greater  diameter;  and 
5:2'.'.  2.02012  :  .S0804  feet,  the  less  diameter  required. 

Or,  the  two  required  diameters  are  24.2414  inches 
and  9.6964  inches. 

Quest.  27.  There  is  a  room  of  wainscot  129  feet  6 
inches  in  circumference,  and  16  feet  9  inches  high  (be- 
ing girt  over  the  mouldings  ;)  there  are  two  windows, 
each  7  feet  3  inches  high,  and  the  breadth  of  each, 
from  cheek  to  cheek,  5  feet  6  inches  ;  the  breadth  of  the 
shutters  of  each  is  4  feet  o  inches.;  the   cheek-boards 


«M-4  Practical  ({urstions. 

and  top  and  bottom-boards  of  each  window,  taken  to- 
gether, are  24  feet  6  inches,  and  their  breadth  1  foot  0 
inches;  the  door-case  7  feet  high,  and  >  feet  B  inches 
uidc:  the  door  8  feet  3  inches  vide  :  I  demand  how 
many  yards  of  wainscot  are  contained  in  that  room  ? 

The  content  of  the  room  2169     1  6 

The  shutters,  at  work  and  half  97   10  6 

The  door,  at  work  and  half  84     1  6 

The  cheek-boards  83     0  o 


The  sum  2J86  10  6 
The  window-lights  and  > 
door-case  deduct  $ 

9)2282     7  6 


253 


Ans.  253  yards  5  feet. 

Quest.  2S.  There  is  a  wall  which  contains  18225  cu- 
bic feet,  and  the  height  is  5  times  the  breadth,  and  the 
length  8  times  the  height ;  what  is  the  length,  breadth, 
and  height  ? 

Assume  the  breadth  1,  then  the  height  must  be  5, 
and  length  40  ;  hence  the  solidity  will  be  200.  Then 
say 

200  :  1,  cube  of  the  breadth  :  :  18223  :  91.125,  the 
cube  of  the  real  breadth ;  the  cube-root  of  which  is  4.5 
the  breadth;  hence  the  height  is  22.5,  and  the  length 
480  feet 


Practical  Questions. 

Quest.  29.  There  is  a  may-pole,  whose 
top-end  was  broken  oft' by  a  blast  of  wind, 
and,  in  falling,  struck  the  ground  at  15  feet 
distance  from  the  bottom  of  the  may -pole  : 
the  broken  pieee  was  39  feet  ;  now  I  de- 
mand the  length  of  the  may -pole  ? 

From  the  square  of  39,  the  length  of  the 
broken  pieee,  subtract  the  square  of  13 ; 
the  square-root  of  the  remainder  is  the 
piece  standing;  to  which  add  the  piece 
broken  oil',  and  you  have  the  whole  length. 
Thus  you  will  find 

The  pieee  standing  to  be  36  feet. 

The  piece  broken  off  is  39  feet. 


The  whole  length  is  73 


&05 

j 

I 

\ 


Quest.  30.     A  may -pole  there  was,    whose  height  I 
would  know, 
The  sun  shining  clear,  strait  to  work  I  did  go. 
The  length  of  the  shadow,  upon  level  ground, 
Just  sixty-five  feet,  when  measurd,  I  found  ; 
AstaffI  had  there.just  five  feet  in  length — 
The  length  of  its  shadow  was  four  feet  one-tenth  : 
How  high  was  the  may -pole,  I  gladly  would  know? 
And  it  is  the  thing  jou're  desired  to  show. 

Here  A  B  represents 
the  length  of  the  shadow 
of  the  may -pole,  andBC 
its  height  :  «A  the  sha- 
dow of  the  stall',  and  A6 
its  height. 

By  similar  triangles. 


ah  :  Ab  :  :  AB  :  BC. 

The    height  BC  will  be 
found  to  be  T9.26  feet. 


/ 

C 

/ 

A 

' 

/ 

/ 

<c 

/ 

<N 

1 

A 

/ 

•y.*y 

6S 

B 

300  Practical  Questions. 

Quest.  M.  What  will  be  the  diameter  of  a  globe, 
when  thr  solidity  aud  superficial  content,  of  it  are 
equal  ? 

•fas.  6:  When  the  diameter  of  the  globe  is  1,  the  so- 
lidity is  to  the  superficies, as  i  to  0. 

Quest.  .12.  What  will  the  axis  of  a  globe  be,  when 
thr  solidity  is  in  proportion  to  the  superficies,  as  IS 
to  8?  JhtS.  13.5.. 

Quest.  33.  There  are  three  grenado  shells,  of  such 
capacity,  that  the  second  shell  will  just  lie  in  the  con- 
cavity of  the  first,  and  the  third  in  the  concavity  of 
the  second.  The  solidity  of  the  metal  of  the  first 
shell  is  equal  to  the  number  of  solid  inches  which 
would  fill  its  capacity  :  the  solidity  of  the  metal  of  the 
second  shell  is  in  proportion  to  the  number  of  solid 
inches  which  would  fill  its  concavity,  as  7  is  to  5  ; 
and  the  solidity  of  the  metal  of  the  third  shell  is  in 
proportion  to  the  number  of  solid  inches  which  would 
till  its  concavity,  as  9  is  to  4.  The  diameter  of  the 
first  shell  is  16  inches;  required  the  diameters  of  the 
second  and  third  shells,  the  thickness  of  each  shell,  and 
also  the  weight,  supposing  a  solid  inch  of  iron  to  weigh 
4  ounces  avoirdupois  ? 

If  the  diameter  of  a  globe  be  1,  its  solidity  will  be 
.5236,  and  globes  are  in  proportion  to  each  other  as  the 
cubes  of  their  diameter  :  hence 

1  :  .5236  : :  cube  of  16=4096  :  2144.6656,  the  half 
of  which  is  1072.3328,  the  solidity  of  the  first  shell, 
and  the  number  of  solid  inches  which  would  fill  its  con- 
cavity; which  being  divided  by  .5236,  and  the  cube- 
root  of  the  quotient  extracted,  will  give  12.699,  the 
internal  diameter  of  the  first  shell ;  hence  its  thick- 
ness is  1.65  inches,  and  weight  268.08  pounds. 

SECONDLY. 

Since  the  second  shell  will  just  fill  the  concavity 
of  the  first,  its    external  diameter  must  be  equal  to 


Practical  Questions.  307 

12.699,  and  the  solidity  of  its  metal  and  concavity  to- 
gether, equal  to  1072.3328  inches. 

745  :  1072.3328  : :  5  :  446.80533,  the  solid  inches 
which  would  fill  the  concavity  of  the  second  shell ; 
which  being  divided  by  .5236,  and  the  cube-root  of  the 
quotient  extracted,  will  give  9.465  inches,  the  internal 
diameter  of  the  second  shell ;  hence  its  thickness  is 
1.607  inches,  and  weight  156.38  pounds. 

THIRDLY. 

Since  the  third  shell  will  just  fill  the  concavity  of 
the  second,  its  external  diameter  must  be  equal  to 
9.485  inches,  and  the  solidity  of  its  metal  and  concav- 
ity together,  equal  to  446.80533  inches. 

9+4  :  446.80533  ::  4  :  137.47846,  the  solid  inches 
which  would  fill  the  concavity  of  the  third  shell  $  which 
being  divided  by  .5236,  and  the  cube-root  of  the  quo- 
tient extracted,  will  give  6.4034  inches,  the  internal 
diameter  of  the  third  shell;  hence  its  thickness  is 
1.541  inch,  and  weight  77.33  pounds. 


FINIS, 


{ 


EXPLANATION  OF  THE   CHARACTERS    MADE    USE    OF  IN 
THE    FOREGOING    "WORK. 

Charact.  Names.  Significations. 

p,  "J  the  sign  of  addition,  as  2-|-4  signi- 

llus,or    (nestnat2and  4  are  to   be  added 
more>    J  together. 
f   \f  1  *ne  s*Sn  °^  subtraction,  as  8 — 3  sig- 

—  i    Minus>    Unifies  that    3    is    to    be   subtracted 
1    orless    J  from  8. 

f  multipli- 1  the  sign  of  multiplication,  as  7X5 
X    i  edinto  or  y  signifies  that  7  is  to  be  multiplied 
(.       DJ>      J  *nto  or  Dy  5* 

f    r   'A  A  1  ine  slSn  °^  division,   as  9~3  signi- 
^    i    uiviaea    (  fieslhat9  isto  be  divided  by  3:and 
(.         ^'       J  I  or  3"95  signifies  the  same. 
_  a  the  sign  of  equality,  as  9—9  signi- 

J     equal     /  fies  that  9  is  equal  to  9,  or  5-f-4 — 
to        f  2=7  signifies  that  5,  increased   by 
*-  }  4-  and  diminished  by  2  is  equal  to  7. 

C  Propor-  )  as  2  :  4  :  :  8  :  1 6  signifies  that  2  is 
:  : :  ;  l      tion,      $  *o  4  as  S  is  to  16. 
w2,  m%  signifies  the  square  or  cube  of  the  letter  m. 

TS2  signifies  the  square  of  the  line  TS. 

y/a\  signifies  the  square  root  of  a  A. 


APPENDIX. 


WEIGHTS  AND  DIMENSIONS 


BALLS  AND  SHELLS. 


THE  weights  of  bodies,  composed  of  the  same  sub- 
stance, are  proportional  to  their  magnitudes;  and  the 
magnitudes  of  similar  bodies  are  proportional  to  the 
cubes  of  their  corresponding  lineal  dimensions  5  there- 
fore, sinee  all  globes  are  similar  bodies,  the  weights  of 
those,  which  are  formed  of  the  same  substance,  must  be 
proportional  to  the  cubes  of  their  diameters.  It  is 
well  known,  however,  that  different  portions  of  metal, 
even  from  the  same  casting,  will  differ  a  small  matter, 
in  density ;  and  also  that  a  still  greater  difference  is 
produced  by  different  degrees  of  temperature :  the 
theory  here  laid  down,  is  therefore  not  to  be  understood 
as  absolutely  true,  but  sufficiently  so  for  the  purposes 
to  which  it  is  applied. 


FEOBLEM    I. 

Given  the  diameter  of  an  Iron  Ball,  to  find  its  weight. 

An  iron  ball  of  4  inches  diameter,  is  known,  from 
experiment,  to  weigh  9  lb.  avoirdupoise ;  therefore,  since 
the  weights  are  as  the  cubes  of  the  diamaters,  it  will 
be,  as  the  cube  of  4,  (or  64)  is  to  9,  so  is  the  cube  of 

Dd 


APPENDIX. 


the  given  diameter,  to  the  weight  required.     Hem* 
^®j  of  the  cube  of  the  diameter  will   be  the  weight  in 

But  ^  =J-|-.i  of  £;  which  gives  the  following 


RULE. 


To  ^  of  the  cube  of  the  diameter,  add  £  of  that  |  £ 


the  sum  will  be  the  weight  in  lbs. 


EXAMPLES. 

1.  Required  the  weight  of  an  iron  ball  whose  diam- 
eter is  6.4  inches. 


The  cube  of  6.1  is         262.14* 


i  of  which  is      -      -       32.768 
|  of  |  is       -     -     -     -       4.096 


Weiffht  of  the  ball     -      36.864  lb. 


2.  The  diameter  of  an  iron  ball  is  8  inches,  what  is 
its  weight.  Ans.  72  lb. 

3.  What  is  the  weight  of  an  iron  ball,  its  diameter 
being  2.4  inches.  Jins.  1  lb.  15.1  oz. 

PROBLEM    II. 

To  find  the  weight  of  a  Leaden  Ball. 

A.  leaden  ball  of  3  inches  diameter  weighs  6  lb. 
Therefore  as  the  cube  of  3  is  to  6  ;  that  is,  as  9  to  2 
so  is  the  cube  of  the  diameter  to  the  weight  of  a 
leaden  ball.     Hence  the  following  practical 


Take  |  of  the  cube  of  the  diameter  of  a  leaden  ball, 
for  its  weight  in  pounds. 


APPENDIX. 

EXAMPLES. 

1.  Required  the  weight  of  a  leaden  ball,  whose  di- 
amter  is  6.3  inches. 

|  of  6.3X6.3X6.3=5.3.366  lb.  the  weight  required. 

2.  What  is  the  weight  of  a  leaden  ball,  8.1  inches 
diameter?  Jlns.  118  lb. 

3.  The  diameter  of  a  leaden  ball  is  1.8  inches,  what 
is  its  weight  ?  Jlns.  1  lb.  4.736  oz. 

PROBLEM    III. 

Given  the  weight  of  an  Iron  Ball,  to  find  its  diameter. 
The  converse  of  prob.  1,  will  give  this 

RULE. 

Multiply  the  given  weight  by  7-|,  and  the  cube  root 
of  the  product  will  be  the  diameter:  or,  to  the  con- 
stant logarithm  0.85194  add  the  logarithm  of  the 
weight  -ifl  pounds,  and  |  of  the  sum  will  be  the  loga- 
rithm of  the  diameter,  in  inches. 

EXAMPLES. 

Required  the  diameter  of  a  42  lb.  iron  ball. 
Constant  log.     0.8/5194 
Los.  of  42  1.62325 


3)2.47519 


Diam.  6.683,  log.    0.82506 


2.  What  is  the  diameter  of  an  iron  ball,  weighing 
18  lb.  Ans.  5.04  inches. 

3.  An  iron  ball  weighs  6  lb.  what  is  its  diameter. 

Jlns,  3.494  inches. 


APPENDIX. 

PROBLEM   IV. 

The  weight  of  a  Leaden  Ball  being  given,  to  find  it* 
diameter. 

The  converse  of  prob.  2,  will  give  this 

RULE. 

Multiply  the  weight  by  4*,  and  the  cube  root  of  the 
product  will  be  the  diameter :  or,  to  the  constant  log- 
arithm 0.65321,  add  the  logarithm  of  the  weight,  and 
|  of  the  sum  will  be  the  logarithm  of  the  diameter. 

EXAMPLES. 

1.  Required  the  diameter  of  a  64  lb.  leaden  ball. 
Constant  log.  0.65321 

Log.  of  64  1.80618 


Diam.  6.605,  log.       3)2.45939  f»  % 

0.81979 

2.  What  is  the  diameter  of  a  leaden  ball  weighing 
18  lbs  ?  Ans.  4.327  inches. 

3.  A  leaden  ball  weighs  9  lbs.  what  is  its  diameter  ? 

Jins.  3.434  inches. 


PROBLEM   V. 

To  find  the  weight  of  an  Iron  Shell. 


To  ^  of  the  difference  of  the  cubes  of  the  external 
and  internal  diameters,  add  its  j,  the  sum  will  be  the 
weight  of  the  shell. 


APPENDIX. 

EXAMPLES. 


1.  Required  the  weight  of  an  iron  shell,  the   exter-#^j 
nal  diameter  being  It,  and  the  internal,  9  inches. 
Tire  cube  of  11  is  1331 

The  cube  of  9  729 


Difference 

602 

|of| 

Hi 

Weight  of  the  shell 

**§1 

2.  To  find   the   weight  of  a  shell   whose  external 
diameter  is  9.6  inches,  and  internal  7.2  inches. 

Jins.  71.928  lb. 

3.  What  is  the  weight  of  an  iron  shell,   the  outer 
diameter  being  18  inches,  and  the  inner  16. 

4g£,  Jins.  244  J  lb. 

PROBLEM    VI. 

To  find  what  weight  of  Powder  will  fill  a  given  ShelL 

>     ^»-       ~V*I%*:"RULE. 

Divide  the  cube  of  the  internal  diameter,  in  inches, 
by  58,  the  quotient  will  be  the  lbs.  of  powder. 


EXAMPLES. 


1.  AVhat  weight  of  powder  will  fill  a  shell,   whose 
internal  diameter  is  9  inches. 

9X9X9—58— 12|4 lb.  the  quantity  required. 

2.  What    quantity  of  powder   will  fill    a  shell   7.2 
inches  in  diameter.  Jins.  6.435  lb. 

3.  Required   the  weight    of  powder  to    fill   a  sh-'ll 
whose  inner  diameter  is  16  inches.  Jins.  70  £  |  lb. 


L 


U'PEXDIX. 
rnoHi.v.N:    vn. 


To  find  the  diameter  of  a  Shell  to  contain  a  given  weight 
of  Fowder. 


Multiply  the  pounds  of  powder  by  58;  the  eube  root 
of  the  product  will  be  the  internal  diameter,  in  inches. 

EXAMPLES. 

1.  To  find  the  diameter  of  a  shell  to  contain  9  lb. 
of  powder. 

58X9=522;  the  cube-root  of  which  is  8.05  inches 
nearly. 

2.  Required  the  diameter  of  a  shell,  to  contain 
47|  lb.  of  powder.  Jlns.  14  inches,  nearly. 

8.  To  find  the  diameter  of  a  shell,  to  contain  27  lb. 
of  powder.  «te.«flMPncnes* 

N.  D.  The  two  last  rules  were  deduced  from  experiments  on 
the  powder  manufactured  for  the  government  of  the  United 
States. 


cSm+Mr/iL^- 


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